Text Book Back Questions and Answers - Chapter 3 Magnetism and Magnetic Effects of Electric Current 12th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Magnetism and Magnetic Effects of Electric Current
Textual Evaluation Solved
Question 1.
The magnetic field at the center $\mathrm{O}$ of the following current loop is-

(a) $\frac{\mu_0 \dot{I}}{4 r} \otimes$
(b) $\frac{\mu_0 I}{4 r} \odot$
(c) $\frac{\mu_0 \mathrm{I}}{2 r} \otimes$
(d) $\frac{\mu_0 I}{2 r} \odot$
Answer:
(a) $\frac{\mu_0 \mathrm{I}}{4 r} \otimes$
Question 2.
An electron moves straight inside a charged parallel plate capacitor of uniform charge density $\sigma$ The time taken by the electron to cross the parallel plate capacitor when the plates of the capacitor are kept under constant magnetic field of induction is-

(a) $\varepsilon_0 \frac{e l \mathrm{~B}}{\sigma}$
(b) $\varepsilon_0 \frac{l \mathrm{~B}}{\sigma l}$
(c) $\varepsilon_0 \frac{l \mathrm{~B}}{e \sigma}$
(d) $\varepsilon_0 \frac{l \mathrm{~B}}{\sigma}$
Answer:
(d) $\varepsilon_0 \frac{l \mathrm{~B}}{\sigma}$
Question 3.
The force experienced by a particle having mass $m$ and charge $q$ accelerated through a potential difference $\mathrm{V}$ when it is kept under perpendicular magnetic field $\vec{B}$ is-
(a) $\sqrt{\frac{2 q^3 \mathrm{BV}}{m}}$
(b) $\sqrt{\frac{q^3 \mathrm{~B}^2 \mathrm{~V}}{2 m}}$
(c) $\sqrt{\frac{2 q^3 \mathrm{~B}^2 \mathrm{~V}}{m}}$
(d) $\sqrt{\frac{2 q^3 \mathrm{BV}}{m^3}}$
Answer:
(c) $\sqrt{\frac{2 q^3 \mathrm{~B}^2 \mathrm{~V}}{m}}$

Question 4.
A circular coil of radius $5 \mathrm{~cm}$ and 50 turns carries a current of 3 ampere. The magnetic dipole moment of the coil is-
(a) $1.0 \mathrm{amp}-\mathrm{m}^2$
(b) $1.2 \mathrm{amp}-\mathrm{m}^2$
(c) $0.5 \mathrm{amp}-\mathrm{m}^2$
(d) $0.8 \mathrm{amp}-\mathrm{m}^2$
Answer:
(b) $1.2 \mathrm{amp}-\mathrm{m}^2$
Question 5.
A thin insulated wire forms a plane spiral of $\mathrm{N}=100$ tight turns carrying a current $1=8 \mathrm{~m} \mathrm{~A}$ (milli ampere). The radii of inside and outside turns are $\mathrm{a}=50 \mathrm{~mm}$ and $\mathrm{b}=100 \mathrm{~mm}$ respectively. The magnetic induction at the center of the spiral is
(a) $5 \mu \mathrm{T}$
(b) $7 \mu \mathrm{T}$
(c) $8 \mu \mathrm{T}$
(d) $10 \mu \mathrm{T}$
Answer:
(b) $7 \mu \mathrm{T}$
Question 6.
Three wires of equal lengths are bent in the form of loops. One of the loops is circle, another is a semicircle and the third one is a square. They are placed in a uniform magnetic field and same electric current is passed through them. Which of the following loop configuration will experience greater torque ?
(a) circle
(b) semi-circle
(c) square
(d) all of them
Answer:
(a) circle
Question 7.
Two identical coils, each with $\mathrm{N}$ turns and radius $\mathrm{R}$ are placed coaxially at a distance $\mathrm{R}$ as shown in the figure. If I is the current passing through the loops in the same direction, then the magnetic field at a point P which is at exactly at $\frac{R}{2}$ distance between two coils is-

(a) $\frac{8 \mathrm{~N} \mu_0 \mathrm{I}}{\sqrt{5} \mathrm{R}}$
(b) $\frac{8 \mathrm{~N} \mu_0 \mathrm{I}}{5^{3 / 2} \mathrm{R}}$
(c) $\frac{8 \mathrm{~N} \mu_0 \mathrm{I}}{5 \mathrm{R}}$
(d) $\frac{4 \mathrm{~N} \mu_0 \mathrm{I}}{\sqrt{5} \mathrm{R}}$
Answer:
(b) $\frac{8 \mathrm{~N} \mu_0 \mathrm{I}}{5^{3 / 2} \mathrm{R}}$
Question 8.
A wire of length 1 carries a current I along the $\mathrm{Y}$ direction and magnetic field is given by $\vec{B}=\frac{\beta}{\sqrt{ } 3}$ $(\hat{i}+\hat{j}+\hat{k})$. The magnitude of Lorentz force acting on the wire is-
(a) $\sqrt{\frac{2}{\sqrt{3}}} \beta \mathrm{I} l$
(b) $\sqrt{\frac{1}{\sqrt{3}}} \beta \mathrm{I} l$
(c) $\sqrt{2} \beta \mathrm{I} l$
(d) $\sqrt{\frac{1}{2}} \beta \mathrm{I} l$
Answer:
(a) $\sqrt{\frac{2}{\sqrt{3}}} \beta \mathrm{I} l$
Question 9.
A bar magnet of length 1 and magnetic moment $M$ is bent in the form of an arc as shown in figure. The new magnetic dipole moment will be- (NEET 2014)

(a) $\mathrm{M}$
(b) $\frac{\pi}{3} \mathrm{M}$
(c) $\frac{2}{\pi} \mathrm{M}$
(d) $\frac{1}{2} \mathrm{M}$
Answer:
(b) $\frac{\pi}{3} \mathrm{M}$
Question 10.
A non-conducting charged ring of charge $\mathrm{q}$, mass $\mathrm{m}$ and radius $\mathrm{r}$ is rotated with constant angular speed $\omega$. Find the ratio of its magnetic moment with angular momentum is
(a) $\frac{q}{m} \mathrm{M}$
(b) $\frac{2 q}{3} \mathrm{M}$
(c) $\frac{q}{2 m} \mathrm{M}$
(d) $\frac{q}{4 m} \mathrm{M}$
Answer:
(c) $\frac{q}{2 m} \mathrm{M}$
Question 11.
The $\mathrm{B}_{\mathrm{H}}$ curve for a ferromagnetic material is shown in the figure. The material is placed inside a long solenoid which contains $1000 \mathrm{turns} / \mathrm{cm}$. The current that should be passed in the solenoid to demagnetize the ferromagnet completely is-

(a) $1.00 \mathrm{~m} \mathrm{~A}$ (milli ampere)
(b) $1.25 \mathrm{~mA}$
(c) $1.50 \mathrm{~mA}$
(d) $1.75 \mathrm{~mA}$
Answer:
(b) $1.25 \mathrm{~mA}$
Question 12.
Two short bar magnets have magnetic moments $1.20 \mathrm{Am}^2$ and $1.00 \mathrm{Am}^2$ respectively. They are kept on a horizontal table parallel to each other with their north poles pointing towards the south. They have a common magnetic equator and are separated by a distance of $20.0 \mathrm{~cm}$. The value of the resultant horizontal magnetic induction at the mid-point $\mathrm{O}$ of the line joining their centers is (Horizontal components of Earth's magnetic induction is $3.6 \times 10^{-5} \mathrm{~Wb} \mathrm{~m}^{-2}$ ) (NSEP 2000-2001)
(a) $3.60 \times 10^{-5} \mathrm{~Wb} \mathrm{~m}^{-1}$
(b) $3.5 \times 10^{-5} \mathrm{~Wb} \mathrm{~m}^{-1}$
(c) $2.56 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-1}$
(d) $2.2 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-1}$
Answer:
(c) $2.56 \times 10^{-4} \mathrm{~Wb} \mathrm{~m}^{-1}$
Question 13.
The vertical component of Earth's magnetic field at a place is equal to the horizontal component. What is the value of angle of dip at this place?
(a) $30^{\circ}$
(b) $45^{\circ}$
(c) $60^{\circ}$
(d) $90^{\circ}$
Answer:
(b) $45^{\circ}$
Question 14.
A flat dielectric disc of radius $\mathrm{R}$ carries an excess charge on its surface. The surface charge density is $\sigma$. The disc rotates about an $\mathrm{a} \times$ is perpendicular to its plane passing through the center with angular velocity

$\omega$. Find the magnitude of the torque on the disc if it is placed in a uniform magnetic field whose strength is $\mathrm{B}$ which is directed perpendicular to the $\mathrm{a} \times$ is of rotation
(a) $\frac{1}{4} \sigma \omega \pi \mathrm{BR}$
(b) $\frac{1}{4} \sigma \omega \pi \mathrm{BR}^2$
(c) $\frac{1}{4} \sigma \omega \pi \mathrm{BR}^3$
(d) $\frac{1}{4} \sigma \omega \pi \mathrm{BR}^4$
Answer:
(d) $\frac{1}{4} \sigma \omega \pi \mathrm{BR}^4$
Question 15.
A simple pendulum with charged bob is oscillating with time period $T$ and let $\theta$ be the angular displacement. If the uniform magnetic field is switched $\mathrm{ON}$ in a direction perpendicular to the plane of oscillation then-
(a) time period will decrease but $\theta$ will remain constant
(b) time period remain constant but $\theta$ will decrease
(c) both $T$ and $\theta$ will remain the same
(d) both $T$ and $\theta$ will decrease
Answer:
(c) both $T$ and $\theta$ will remain the same
Short Answer Questions
Question 1.
What is meant by magnetic induction?
Answer:
The magnetic induction (total magnetic field) inside the specimen $\vec{B}$ is equal to the sum of the magnetic field $\vec{B}_0$ produced in vacuum due to the magnetising field and the magnetic field $\vec{B}_{\mathrm{m}}$ due to the induced magnetisation of the substance.
$
\vec{B}=\vec{B}_0+\vec{B}_{\mathrm{m}}=\mu_0 \vec{H}+\mu_0 \vec{I}=\mu_0(\vec{H}+\vec{I})=(\vec{H}+\vec{I})
$
Question 2.
Define magnetic flux.
Answer:
The number of magnetic field lines crossing per unit area is called magnetic flux $B$.
$
\Phi_{\mathrm{B}}=\vec{B} \cdot \vec{A}=\mathrm{BA} \cos \theta=\mathrm{B} \perp \mathrm{A}
$
Question 3.
Define magnetic dipole moment.
Answer:
The magnetic dipole moment is defined as the product of its pole strength and magnetic length.
$
\vec{P}=\mathrm{q}_{\mathrm{m}} \vec{d}
$

Question 4.
State Coulomb's inverse law.
Answer:
The force of attraction or repulsion between two magnetic poles is directly proportional to the product of
their pole strengths and inversely proportional to the square of the distance between them.
$
\overrightarrow{\mathrm{F}} \propto \frac{q_{m_{\Lambda}} q_{m_b}}{r^2} \hat{r}
$
Question 5.
What is magnetic susceptibility?
Answer:
It is defined as the ratio of the intensity of magnetisation $(\vec{M})$ induced in the material due to the magnetising field $(\vec{H})$
$
\chi_m=\left|\frac{\vec{M}}{\overrightarrow{\mathrm{H}}}\right|
$
Question 6.
State Biot-Savart's law.
Answer:
The magnitude of magnetic field $\mathrm{d} \vec{B}$ at a point $P$ at a distance $r$ from the small elemental length taken on a conductor carrying current varies-
(i) directly as the strength of the current I
(ii) directly as the magnitude of the length element $\overrightarrow{d l}$.
(iii) directly as the sine of the angle (say, $\theta$ ) between $\vec{d} l$ and $\hat{r}$
(iv) inversely as the square of the distance between the point $P$ and length element $\overrightarrow{d l}$ This is expressed as $d \mathrm{~B} \propto \frac{\mathrm{I} d l}{r^2} \sin \theta$

Question 7.
What is magnetic permeability?
Answer:
Magnetic permeability:
The magnetic permeability can be defined as the measure of ability of the material to allow the passage of magnetic field lines through it or measure of the capacity of the substance to take magnetisation or the degree of penetration of magnetic field through the substance.
Question 8.
State Ampere's circuital law.
Answer:
The line integral of magnetic field over a closed loop is $\mu_0$ times net current enclosed by the loop.
$
\oint_{\mathrm{C}} \overrightarrow{\mathrm{B}} \cdot d l=\mu_0 \mathrm{I}_{\text {enciosed }}
$
Question 9.
Compare dia, para and ferromagnetism.
Answer:

Question 10.
What is meant by hysteresis?
Answer:
The phenomenon of lagging of magnetic induction behind the magnetising field is called hysteresis.
Hysteresis means Tagging behind'.
Long Answer Questions
Question 1.
Discuss Earth's magnetic field in detail.
Answer:
Gover suggested that the Earth's magnetic field is due to hot rays coming out from the Sun. These rays will heat up the air near equatorial region. Once air becomes hotter, it rises above and will move towards northern and southern hemispheres and get electrified. This may be responsible to magnetize the ferromagnetic materials near the Earth's surface.

The north pole of magnetic compass needle is attracted towards the magnetic south pole of the Earth which is near the geographic north pole. Similarly, the south pole of magnetic compass needle is attracted towards the geographic north pole of the Earth which is near magnetic north pole. There are three quantities required to specify the magnetic field of the Earth on its surface, which are often called as the elements of the Earth's magnetic field. They are:
(a) Magnetic declination (D):
The angle between magnetic meridian at a point and geographical meridian is called the declination or magnetic declination (D).
(b) Magnetic dip or inclination (I):
The angle subtended by the Earth's total magnetic field $\vec{B}$ with the horizontal direction in the magnetic meridian is called dip or magnetic inclination (I) at that point.
(c) The horizontal component of the Earth's magnetic field $\left(\mathrm{B}_{\mathrm{H}}\right)$ :
The component of Earth's magnetic field along the horizontal direction in the magnetic meridian is called horizontal component of Earth's magnetic field, denoted by $\mathrm{B}_{\mathrm{H}}$.
Let $\mathrm{B}_{\mathrm{E}}$ be the net Earth's magnetic field at a point $\mathrm{P}$ on the surface of the Earth. $\mathrm{B}_{\mathrm{E}}$ can be resolved into two perpendicular components.
Horizontal component $\mathrm{B}_{\mathrm{H}}=\mathrm{B}_{\mathrm{E}} \cos \mathrm{I}$
Vertical component $\mathrm{B}_{\mathrm{V}}=\mathrm{B}_{\mathrm{E}} \sin \mathrm{I} \ldots \ldots$ (2)
Dividing the equation, we get $\tan \mathrm{I}=\frac{B_V}{B_H}$
(i) At magnetic equator The Earth's magnetic field is parallel to the surface of the Earth (i.e., horizontal) which implies that the needle of magnetic compass rests horizontally at an angle of dip, $I=0^{\circ}$.
$\mathrm{B} \mathrm{B}_{\mathrm{E}}=\mathrm{B}_{\mathrm{E}}$
$\mathrm{Bv}=0$
This implies that the horizontal component is maximum at equator and vertical component is zero at equator.
(ii) At magnetic poles. The Earth's magnetic field is perpendicular to the surface of the Earth (i.e., vertical) which implies that the needle of magnetic compass rests vertically at an angle of dip, $I=90^{\circ}$ Hence, $\mathrm{B}_{\mathrm{H}}=0$
$
\mathrm{Bv}=\mathrm{B}_{\mathrm{E}}
$
This implies that the vertical component is maximum at poles and horizontal component is zero at poles.

Question 2.
Deduce the relation for the magnetic induction at a point due to an infinitely long straight conductor carrying current.
Answer:
Magnetic field due to long straight conductor carrying current:
Consider a long straight wire $N M$ with current I flowing from $N$ to $M$. Let $P$ be the point at a distance a from point $\mathrm{O}$. Consider an element of length $\mathrm{dl}$ of the wire at a distance 1 from point $\mathrm{O}$ and $\vec{r}$ be the vector joining the element $\mathrm{dl}$ with the point $P$. Let $\theta$ be the angle between $\overrightarrow{d l}$ and $\vec{r}$. Then, the magnetic field at $\mathrm{P}$ due to the element is $\mathrm{d} \vec{B}=\frac{\mu_0 I \overrightarrow{d l}}{4 \pi r^2} \sin \theta$ (unit vector perpendicular to $\overrightarrow{d l}$ and $\vec{r}$ ) ......(1)
The direction of the field is perpendicular to the plane of the paper and going into it. This can be determined by
taking the cross product between two vectors $\overrightarrow{d l}$ and $\vec{r}$ (let it be $\hat{n}$ ). The net magnetic field can be determined by integrating equation with proper limits.

$
\vec{B}=\int d \vec{B}
$
From the figure, in a right angle triangle $P A O$, $\tan (\pi-\theta)=\frac{a}{l}$
$1=\frac{a}{\tan \theta}($ since $\tan (\pi-\theta)=-\tan \theta) \Rightarrow \frac{1}{\tan \theta}$
$1=a \cot \theta$ and $r=a \operatorname{cosec} \theta$
differentiating,
$\mathrm{dl}=\mathrm{a} \operatorname{cosec}^2 \theta \mathrm{d} \theta$

$\begin{aligned}
& d \overrightarrow{\mathrm{B}}=\frac{\mu_0 \mathrm{I}}{4 \pi} \frac{\left(a \operatorname{cosec}^2 \theta d \theta\right)}{(a \operatorname{cosec} \theta)^2} \sin \theta d \theta \hat{n} \\
& =\frac{\mu_0 I}{4 \pi} \frac{\left(a \operatorname{cosec}^2 \theta d \theta\right)}{a^2 \operatorname{cosec}^2 \theta} \sin \theta d \theta \hat{n} \\
& d \overrightarrow{\mathrm{B}}=\frac{\mu_0 \mathrm{I}}{4 \pi a} \sin \theta d \theta \hat{n} \\
&
\end{aligned}$

This is the magnetic field at a point $P$ due to the current in small elemental length. Note that we have expressed the magnetic field OP in terms of angular coordinate i.e. $\theta$. Therefore, the net magnetic field at the point $P$ which can be obtained by integrating $\mathrm{d} \vec{B}$ by varying the angle from $\theta=\left(\varphi_1\right.$ to $\theta=\varphi_1$ is $\overrightarrow{\mathrm{B}}=\frac{\mu_0 \mathrm{I}}{4 \pi a} \int_{\varphi_1}^{\varphi_2} \sin \theta d \theta \hat{n}=\frac{\mu_0 I}{4 \pi a}\left(\cos \varphi_1-\cos \varphi_2\right) \hat{n}$
For a an infinitely long straight wire, $1=0$ and $2=$, the magnetic field is $\vec{B}=\frac{\mu_0}{2 \pi a} \hat{n}$
Note that here $\hat{n}$ represents the unit vector from the point $\mathrm{O}$ to $\mathrm{P}$.
Question 3.
Obtain a relation for the magnetic induction at a point along the axis of a circular coil carrying current.
Answer:
Magnetic field produced along the axis of the current carrying circular coil:
Consider a current carrying circular loop of radius $\mathrm{R}$ and let $\mathrm{I}$ be the current flowing through the wire in the direction. The magnetic field at a point $\mathrm{P}$ on the axis of the circular coil at a distance $\mathrm{z}$ from its center of the coil O. It is computed by taking two diametrically opposite line elements of the coil each of length $\overrightarrow{d l}$ at $\mathrm{C}$ and $\mathrm{D}$. Let $\vec{r}$ be the vector joining the current element $(1 \overrightarrow{d l})$ at $\mathrm{C}$ to the point $\mathrm{P}$.
$\mathrm{PC}=\mathrm{PD}=\mathrm{T}=\sqrt{R^2+Z^2}$ and
angle $\angle \mathrm{CPO}=\angle \mathrm{DPO}=\theta$
According to Biot-Savart's law, the magnetic field at $\mathrm{P}$ due to the current element I $\overrightarrow{d l}$ is $\mathrm{d} \vec{B}=\frac{\mu_0}{4 \pi} \frac{\mathrm{I} d \vec{l} \times \hat{r}}{r^2} \ldots .(1)$
The magnitude of magnetic field due to current element $1 \mathrm{dl}$ at $\mathrm{C}$ and $\mathrm{D}$ are equal because of equal distance from the coil. The magnetic field $\mathrm{dB}$ due to each current element $\mathrm{I} \overrightarrow{d l}$ is resolved into two components; $\mathrm{dB} \sin \theta$ along $\mathrm{y}$-direction and $\mathrm{dB} \cos \theta$ along $\mathrm{z}$-direction. Horizontal components of each current element cancels out while the vertical components $(\mathrm{dB} \cos \theta \hat{k})$ alone contribute to total magnetic field at the point $P$.

If we integrate $\overrightarrow{d l}$ around the loop, $\mathrm{d} \vec{B}$ sweeps out a cone, then the net magnetic field $\vec{B}$ at point $\mathrm{P}$ is
$
\begin{aligned}
& \overrightarrow{\mathrm{B}}=\int d \overrightarrow{\mathrm{B}}=\int d \mathrm{~B} \cos \theta \hat{k} \\
& \overrightarrow{\mathrm{B}}=\frac{\mu_0 \mathrm{I}}{4 \pi} \int \frac{d l}{r^2} \cos \theta \hat{k}
\end{aligned}
$
But $\cos \theta=\frac{\mathrm{R}}{\left(\mathrm{R}^2+\mathrm{Z}^2\right)^{\frac{1}{2}}}$
Using Pythagorous theorem $r^2=R^2+Z^2$ and integrating line element from 0 to $2 \pi R$, we get
$
\overrightarrow{\mathrm{B}}=\frac{\mu_0 \mathrm{I}}{2 \pi} \frac{\mathrm{R}^2}{\left(\mathrm{R}^2+\mathrm{Z}^2\right)^{\frac{3}{2}}} \hat{k}
$
Note that the magnetic field $\vec{B}$ points along the direction from the point $O$ to $P$. Suppose if the current flows in clockwise direction, then magnetic field points in the direction from the point $\mathrm{P}$ to $\mathrm{O}$.
Question 4.
Compute the torque experienced by a magnetic needle in a uniform magnetic field.
Answer:
Torque Acting on a Bar Magnet in Uniform Magnetic Field:
Consider a magnet of length 21 of pole strength $\mathrm{q}_{\mathrm{m}}$ kept in a uniform magnetic field $\vec{B}$ Each pole experiences a force of magnitude $\mathrm{q}_{\mathrm{m}} \mathrm{B}$ but acts in opposite direction.
Therefore, the net force exerted on the magnet is zero, so that there is no translatory motion. These two forces constitute a couple (about midpoint of bar magnet) which will rotate and try to align in the direction of the magnetic field $\vec{B}$.
The force experienced by north pole, $\vec{F}_{\mathrm{N}}=\mathrm{q}_{\mathrm{m}} \vec{B}$
The force experienced by south pole, $\vec{F}_{\mathrm{S}}=\mathrm{q}_{\mathrm{m}} \vec{B}$
Adding equations (1) and (2), we get the net force acting on the dipole as
$
\vec{F}=\vec{F}_{\mathrm{N}}+\vec{F}_{\mathrm{S}}=\vec{O}
$

This implies, that the net force acting on the dipole is zero, but forms a couple which tends to rotate the bar magnet clockwise (here) in order to align it along $\vec{B}$.
The moment of force or torque experienced by north and south pole about point $\mathrm{O}$ is
$
\begin{aligned}
& \vec{\tau}=\overrightarrow{\mathrm{ON}} \times \overrightarrow{\mathrm{F}}_{\mathrm{N}}+\overrightarrow{\mathrm{OS}} \times \overrightarrow{\mathrm{F}}_{\mathrm{S}} \\
& \vec{\tau}=\overrightarrow{\mathrm{ON}} \times q_m \overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{OS}} \times\left(-q_m \overrightarrow{\mathrm{B}}\right)
\end{aligned}
$
By using right hand cork screw rule, we conclude that the total torque is pointing into the paper. Since the magnitudes
$
|\overrightarrow{\mathrm{ON}}|=|\overrightarrow{\mathrm{OS}}|=l \text { and }\left|q_m \overrightarrow{\mathrm{B}}\right|=\left|-q_m \overrightarrow{\mathrm{B}}\right|
$
The magnitude of total torque about point $\mathrm{O}$
$
\begin{aligned}
& \tau=1 \times \mathrm{q}_{\mathrm{m}} \mathrm{B} \sin \theta+1 \times \mathrm{q}_{\mathrm{m}} \mathrm{B} \sin \theta \\
& \tau=21 \times \mathrm{q}_{\mathrm{m}} \mathrm{B} \sin \theta \\
& \tau=\mathrm{P}_{\mathrm{m}} \mathrm{B} \sin \theta \\
& \left(\therefore \mathrm{q}_{\mathrm{m}} \times 21=\mathrm{P}_{\mathrm{m}}\right)
\end{aligned}
$
In vector notation, $\tau=\mathrm{p}_{\mathrm{m}} \times \vec{B}$
Question 5.
Calculate the magnetic induction at a point on the $a \times i a l$ line of a bar magnet.
Answer:
Magnetic field at a point along the axial line of the magnetic dipole (bar magnet):
Consider a bar magnet NS. Let N be the North Pole and $\mathrm{S}$ be the south pole of the bar magnet, each of pole strength $\mathrm{q}_{\mathrm{m}}$ and separated by a distance of 21. The magnetic field at a point $\mathrm{C}$ (lies along the axis of the magnet) at a distance from the geometrical center $\mathrm{O}$ of the bar magnet can be computed by keeping unit north pole $\left(\mathrm{q}_{\mathrm{MC}}=1 \mathrm{~A} \mathrm{~m}\right)$ at $\mathrm{C}$. The force experienced by the unit north pole at $\mathrm{C}$ due to pole strength can be computed using Coulomb's law of magnetism as follows:
The force of repulsion between north pole of the bar magnet and unit north pole at point $\mathrm{C}$ (in free space) is
$
\overrightarrow{\mathrm{F}}_{\mathrm{N}}=\frac{\mu_0}{4 \pi} \frac{q_m}{(r-l)^2} \hat{l}
$
where $\mathrm{r}-1$ is the distance between north pole of the bar magnet and unit north pole at $\mathrm{C}$. The force of attraction between South Pole of the bar magnet and unit North Pole at point C (in free space) is

$
\overrightarrow{\mathrm{F}}_{\mathrm{S}}=-\frac{\mu_0}{4 \pi} \frac{q_m}{(r+l)^2} \hat{l}
$
where $r+1$ is the distance between south pole of the bar magnet and unit north pole at $\mathrm{C}$.

From equation (1) and (2), the net force at point $\mathrm{C}$ is $\vec{F}=\vec{F}_{\mathrm{N}}+\vec{F}_{\mathrm{S}}$.
From definition, this net force is the magnetic field due to magnetic dipole at a point $\mathrm{C}(\vec{F}=\vec{B})$
$
\begin{aligned}
& \overrightarrow{\mathrm{B}}=\frac{\mu_0}{4 \pi(r-l)^2} \hat{i}+\left(\frac{\mu_0}{4 \pi} \frac{q_m}{(4+l)^2} \hat{i}\right)=\frac{\mu_0 q_m}{4 \pi}\left(\frac{1}{(r-l)^2}-\frac{1}{(r+l)^2}\right) \hat{i} \\
& \overrightarrow{\mathrm{B}}=\frac{\mu_0 2 r}{4 \pi}\left(\frac{q_m \cdot(2 l)}{\left(r^2-l^2\right)^2}\right) \hat{i}
\end{aligned}
$
Since, magnitude of magnetic dipole moment is $\left|\vec{P}_m\right| \mathrm{p}_{\mathrm{m}}=\mathrm{q}_{\mathrm{m}}$. 21 the magnetic field point C equation (3) can be written as
$
\overrightarrow{\mathrm{B}}_{\text {axial }}=\frac{\mu_0}{4 \pi}\left(\frac{2 r p_m}{\left(r^2-l^2\right)^2}\right) \hat{i}
$
If the distance between two poles in a bar magnet are small (looks like short magnet) compared to the distance between geometrical centre $\mathrm{O}$ of bar magnet and the location of point $\mathrm{C}$ i.e.,
$r \gg 1$ then, $\left(r^2-1^2\right)^2 \approx r^4 \ldots(5)$
Therefore, using equation (5) in equation (4), we get
$\overrightarrow{\mathrm{B}}_{\mathrm{axial}}=\frac{\mu_0}{4 \pi}\left(\frac{2 r p_m}{\left(r^3\right.}\right) \hat{i}=\frac{\mu_0}{4 \pi} \frac{2}{r^3} \vec{p}_m$
Where $\vec{p}_m=p_m \hat{i}$.
Question 6.
Obtain the magnetic induction at a point on the equatorial line of a bar magnet. Magnetic field at a point along the equatorial line due to a magnetic dipole (bar magnet)
Answer:
Consider a bar magnet NS. Let $\mathrm{N}$ be the north pole and $\mathrm{S}$ be the south pole of the bar magnet, each with pole strength $\mathrm{q}_{\mathrm{m}}$ and separated by a distance of 21 . The magnetic field at a point $\mathrm{C}$ (lies along the equatorial line) at a distance $\mathrm{r}$ from the geometrical center $\mathrm{O}$ of the bar magnet can be computed by keeping unit north pole $\left(\mathrm{q}_{\mathrm{m}} \mathrm{C}=1 \mathrm{~A} \mathrm{~m}\right)$ at $\mathrm{C}$. The force experienced by the unit north pole at $\mathrm{C}$ due to pole strength N-S can be computed using Coulomb's law of magnetism as follow's:
The force of repulsion between North Pole of the bar magnet and unit north pole at point C (in free space) is

$
\vec{F}_{\mathrm{N}}=-\mathrm{F}_{\mathrm{N}} \cos \theta \hat{i}+\mathrm{F}_{\mathrm{N}} \sin \theta \hat{j}
$
Where $\mathrm{F}_{\mathrm{N}}=\frac{\mu_0}{4 \pi} \frac{q_m}{r^{r^2}}$
The force of attraction (in free space) between south pole of the bar magnet and unit north pole at point C is
$
\vec{F}_{\mathrm{S}}=-\mathrm{F}_{\mathrm{S}} \cos \theta \hat{i}+\mathrm{F}_{\mathrm{S}} \sin \theta \mathrm{j}
$
Where $\vec{F}_{\mathrm{S}}=\frac{\mu_0}{4 \pi} \frac{q_m}{r^2}$
From equation (1) and equation (2), the net force at point $\mathrm{C}$ is $\vec{F}=\mathrm{F}_{\mathrm{N}}+\mathrm{F}_{\mathrm{S}}$ This net force is equal to the magnetic field at the point $\mathrm{C}$.
$
\vec{B}=-\left(\mathrm{F}_{\mathrm{N}}+\mathrm{F}_{\mathrm{S}}\right) \cos \theta \hat{i}
$
Since, $\mathrm{F}_{\mathrm{N}}=\mathrm{F}_{\mathrm{S}}$

$
\overrightarrow{\mathrm{B}}=-\frac{2 \mu_0}{4 \pi} \frac{q_m}{r^{\prime 2}} \cos \theta \hat{i}=-\frac{2 \mu_0}{4 \pi} \frac{q_m}{\left(r^2+l^2\right)}-\cos \theta \hat{i}
$
In a right angle triangle $\mathrm{NOC}$ as shown in the Figure 1
$
\cos \theta=\frac{\text { adjacent }}{\text { hypotenuse }}=\frac{1}{r^{\prime}}=\frac{1}{\left(r^2+l^2\right)^{\frac{1}{2}}}
$
Substituting equation 4 in equation 3 We get
$
\overrightarrow{\mathrm{B}}=-\frac{\mu_0}{4 \pi} \frac{q_m \times(2 l)}{\left(r^2+l^2\right)^{\frac{3}{2}}}
$
Since, magnitude of magnetic dipole moment is $\left|\vec{P}_m\right|=\mathrm{P}_{\mathrm{m}}=\mathrm{q}_{\mathrm{m}} .21$ and substituting in equation (5), the magnetic field at a point $\mathrm{C}$ is
$
\overrightarrow{\mathrm{B}}_{\text {equatorial }}=-\frac{\mu_0}{4 \pi} \frac{\mathrm{P}_m}{\left(r^2+l^2\right)^{\frac{3}{2}}} \hat{i}
$
If the distance between two poles in a bar magnet are small (looks like short magnet) when compared to the distance between geometrical center $\mathrm{O}$ of bar magnet and the location of point $\mathrm{C}$ i.e., $\mathrm{r}>>1$, then, $\left(r^2+1^2\right)^{\frac{3}{2}}$
Therefore, using equation (7) in equation (6), we get $\approx r^3$
$\vec{B}_{\text {equatorial }}=-\frac{\mu_0}{4 \pi} \frac{p_m}{r^3} \hat{i}$
Since $\mathrm{P}_{\mathrm{m}} \hat{i}=\left|\vec{P}_m\right|_{\mathrm{m}}$, in general, the magnetic field at equatorial point is given by
$\overrightarrow{\mathrm{B}}_{\text {equatorial }}=-\frac{\mu_0}{4 \pi} \frac{p_m}{r^3} \ldots(8)$
Note that magnitude of $\mathrm{B}_{\text {axial }}$ is twice that of magnitude of $\mathrm{B}_{\text {equatorial }}$ and the direction of are opposite.

Question 7.
Find the magnetic induction due to a long straight conductor using Ampere's circuital law. Magnetic field due to the current carrying wire of infinite
Answer:
length using Ampere's law:
Consider a straight conductor of infinite length carrying current $I$ and the direction of magnetic field lines.
Since the wire is geometrically cylindrical in shape $\mathrm{C}$ and symmetrical about its axis, we construct an
Amperian loop in the form of a circular shape at a distance $\mathrm{r}$ from the centre of the conductor. From the Ampere's law, we get

$
\oint_{\mathrm{C}} \overrightarrow{\mathrm{B}} \cdot d \vec{l}=\mu_0 \mathrm{I}
$
Where $\mathrm{dl}$ is the line element along the amperian loop (tangent to the circular loop). Hence, the angle between magnetic field vector and line element is zero. Therefore,
$
\oint_{\mathrm{C}} \mathrm{B} \cdot d l=\mu_0 \mathrm{I}
$
where I is the current enclosed by the Amperian loop. Due to the symmetry, the magnitude of the magnetic field is uniform over the Amperian loop, we can take B out of the integration.
$
\mathrm{B} \oint_{\mathrm{C}} d l=\mu_0 \mathrm{I}
$
For a circular loop, the circumference is $2 \pi \mathrm{r}$, which implies,
$
\mathrm{B} \int_{\hat{n}}^{2 \pi r} d l=\mu_0 I
$
$\overrightarrow{\mathrm{B}} .2 \pi r=\mu_0 \mathrm{I} \Rightarrow \overrightarrow{\mathrm{B}}=\frac{\mu_0 \mathrm{I}}{2 \pi r}$
In vector form, the magnetic field is
$
\vec{B}=\frac{\mu_0 I}{2 \pi r} \hat{n}
$
Where $\hat{n}$ is the unit vector along the tangent to the Amperian loop. This perfectly agrees with the result obtained from Biot-Savarf s law as given in equation
$
\vec{B}=\frac{\mu_0 I}{2 \pi a} \hat{n}
$

Question 8.
Discuss the working of cyclotron in detail.
Answer:
Cyclotron:
Cyclotron is a device used to accelerate the charged particles to gain large kinetic energy. It is also called as high energy accelerator. It was invented by Lawrence and Livingston in 1934.
Principle:
When a charged particle moves normal to the magnetic field, it experiences magnetic Lorentz force.
Construction:
The particles are allowed to move in between two semicircular metal containers called Dees (hollow D shaped objects). Dees are enclosed in an evacuated chamber and it is kept in a region with uniform magnetic field controlled by an electromagnet. The direction of magnetic field is normal to the plane of the Dees. The two Dees are kept separated with a gap and the source $S$ (which ejects the particle to be accelerated) is placed at the center in the gap between the Dees. Dees are connected to high frequency alternating potential difference.

Working:
Let us assume that the ion ejected from source $\mathrm{S}$ is positively charged. As soon as ion is ejected, it is accelerated towards a Dee (say, Dee - 1) which has negative potential at that time. Since the magnetic field is normal to the plane of the Dees, the ion undergoes circular path. After one semi-circular path in Dee-1, the ion reaches the gap between Dees. At this time, the polarities of the Dees are reversed so that the ion is now accelerated towards Dee-2 with a greater velocity. For this circular motion, the centripetal force of the charged particle $q$ is provided by Lorentz force.
$
\frac{m v^2}{r} \mathrm{qvB} \Rightarrow \mathrm{r}=\frac{m}{q b} \mathrm{v} \Rightarrow \mathrm{r} \propto \mathrm{v}
$
From the equation, the increase in velocity increases the radius of circular path. This process continues and hence the particle undergoes spiral path of increasing radius. Once it reaches near the edge, it is taken out with the help of deflector plate and allowed to hit the target $T$. Very important condition in cyclotron operation is the resonance condition. It happens when the frequency $f$ at which the positive ion circulates in the magnetic field must be equal to the constant frequency of the electrical oscillator $\mathrm{f}_{\mathrm{osc}}$ From equation
$
\mathrm{f}_{\mathrm{osc}}=\frac{q B}{2 \pi m} \mathrm{~T}=\frac{1}{f_{\text {osc }}}
$
The time period of oscillation is
$
\mathrm{T}=\frac{2 \pi m}{q B}
$
The kinetic energy of the charged paricle is
$
\mathrm{KE}=\frac{1}{2} \mathrm{mv}^2=\frac{q^2}{B^2} r^2 2 m
$
Limitations of cyclotron:
(a) the speed of the ion is limited

(b) electron cannot be accelerated
(c) uncharged particles cannot be accelerated
Question 9.
What is tangent law? Discuss in detail.
Answer:
Tangent law:
Statement:
When a magnetic needle or magnet is freely suspended in two mutually perpendicular uniform magnetic fields, it will come to rest in the direction of the resultant of the two fields.
Explanation:
Let B be the magnetic field produced by passing current through the coil of the tangent Galvanometer and $\mathrm{B}_{\mathrm{H}}$ be the horizontal component of earth's magnetic field. Under the action of two magnetic fields, the needle comes to rest making angle $\theta$ with $\mathrm{B}_{\mathrm{H}}$, such that 

$\tan \theta=\frac{B}{B_H}$
$\mathrm{B}=\mathrm{B}_{\mathrm{H}} \tan \theta$
When no current is passed through the coil, the small magnetic needle lies along horizontal component of Earth's magnetic field. When the circuit is switched ON, the electric current will pass through the circular coil and produce magnetic field. Now there are two fields which are acting mutually perpendicular to each other. They are:
(1) the magnetic field (B) due to the electric current in the coil acting normal to the plane of the coil.
(2) the horizontal component of Earth's magnetic field $\left(\mathrm{B}_{\mathrm{H}}\right)$
Because of these crossed fields, the pivoted magnetic needle deflects through an angle $\theta$. From tangent law, $\mathrm{B}=\mathrm{B}_{\mathrm{H}} \tan \theta$ When an electric current is passed through a circular coil of radius $\mathrm{R}$ having $\mathrm{N}$ turns, the magnitude of magnetic field at the center is
$
\mathrm{B}=\mu_0 \frac{N I}{2 R} \cdots \cdots(2)
$
From equation (1) and equation (2), we get
$
\mu_0 \frac{N I}{2 R}=\mathrm{B}_{\mathrm{H}} \tan \theta
$
The horizontal component of Earth's magnetic field can be determined as
$
\mathrm{B}=\mu_0 \frac{N I}{2 \operatorname{Rtan} \theta} \text { in tesla }
$
Question 10.
Explain the principle and working of a moving coil galvanometer.
Answer:
Moving coil galvanometer:
Moving coil galvanometer is a device which is used to indicate the flow of current in an electrical circuit.
Principle:
When a current carrying loop is placed in a uniform magnetic field it experiences a torque.
Construction:
A moving coil galvanometer consists of a rectangular coil PQRS of insulated thin copper wire. The coil contains a large number of turns wound over a light metallic frame. A cylindrical soft-iron core is placed symmetrically inside the coil. The rectangular coil is suspended freely between two pole pieces of a horse-shoe magnet.
The upper end of the rectangular coil is attached to one end of fine strip of phosphor bronze and the lower end of the coil is connected to a hair spring which is also made up of phosphor bronze. deflection of the coil with the help of lamp and scale arrangement. The other end of the mirror is connected to a torsion head $T$. In order to pass electric current through the galvanometer, the suspension strip W and the spring S are connected to terminals.

Working:
Consider a single turn of the rectangular coil $P Q R S$ whose length be 1 and breadth $b . P Q=R S=1$ and $Q R$ $=\mathrm{SP}=\mathrm{b}$. Let $\mathrm{I}$ be the electric current flowing through the rectangular coil PQRS. The horse-shoe magnet has hemi - spherical magnetic poles which produces a radial magnetic field.

Force downwards
Force acting on current carrying coil
Due to this radial field, the sides $\mathrm{QR}$ and $\mathrm{SP}$ are always parallel to to the B-field (magnetic field) and experience no force. The sides $P Q$ and RS are always parallel to the B-field and experience force and due to this, torque is produced.
For single turn, the deflection couple.
$\tau=b \mathrm{bF}=\mathrm{bBI}=(\mathrm{lb}) \mathrm{BI}=\mathrm{ABI}$ since, area of the coil
$A=1 b$ For coil with $N$ turns, we get $r=N A B I \ldots \ldots$ (1)
Due to this deflecting torque, the coil gets twisted and restoring torque (also known as restoring couple) is developed. Hence the magnitude of restoring couple is proportional to the amount of twist $\theta$. Thus $\tau=\mathrm{K} \theta$
( $(2)$
where $\mathrm{K}$ is the restoring couple per unit twist or torsional constant of the spring. At equilibrium, the deflection couple is equal to the restoring couple. Therefore by comparing equation (1) and (2), we get
$\mathrm{NABI}=\mathrm{K} \theta \Rightarrow \mathrm{I}=\frac{K}{N A B} \theta$
(or) $\mathrm{I}=\mathrm{G} \theta$
Where, $\mathrm{G}=\frac{K}{N A B}$ is called is called galvanometer constant or current reduction factor of the galvanometer. Since, suspended moving coil galvanometer is very sensitive, we have to handle with high care while doing experiments. Most of the galvanometer we use arc pointer type moving coil galvanometer.
Question 11.
Discuss the conversion of galvanometer into an ammeter and also a voltmeter.
Answer:
Conversion of galvanometer into ammeter and voltmeter:
A galvanometer is very sensitive instrument to detect the current. It can be easily converted into ammeter and voltmeter.
(i) Galvanometer to an Ammeter:
Ammeter is an instrument used to measure current flowing in the electrical circuit. The ammeter must offer low resistance such that it will not change the current passing through it. So ammeter is connected in series to measure the circuit current.

A galvanometer is converted into an ammeter by connecting a low resistance in parallel with the galvanometer. This low resistance is called shunt resistance $\mathrm{S}$. The scale is now calibrated in ampere and the range of ammeter depends on the values of the shunt resistance.

Let I be the current passing through the circuit. When current I reaches the junction $\mathrm{A}$, it divides into two components. Let $I_g$ be the current passing through the galvanometer of resistance $R_g$ through a path $\mathrm{AGE}$ and the remaining current $\left(\mathrm{I}-\mathrm{I}_{\mathrm{g}}\right)$ passes along the path $\mathrm{ACDE}$ through shunt resistance $\mathrm{S}$.
The value of shunt resistance is so adjusted that current I produces full scale deflection in the galvanometer. The potential difference across galvanometer is same as the potential difference across shunt resistance.
$
\begin{aligned}
& \mathrm{V}_{\text {galvanometer }}=\mathrm{V}_{\text {shunt }} \Rightarrow \mathrm{I}_g \mathrm{R}_g=\left(\mathrm{I}-\mathrm{I}_g\right) \mathrm{S} \\
& \mathrm{S}=\frac{\mathrm{I}_s}{\left(\mathrm{I}-\mathrm{I}_g\right)} \mathrm{R}_g \quad \text { (or) } \quad \mathrm{I}_g=\frac{\mathrm{S}}{\mathrm{S}+\mathrm{R}_g} \mathrm{l} \Rightarrow \mathrm{I}_g \propto \mathrm{I}
\end{aligned}
$
Since, the deflection in the galvanometer is proportional to the current passing through it.
$
\theta=\frac{1}{g} \mathrm{I}_{\mathrm{g}} \Rightarrow \theta \mathrm{Ig} \Rightarrow \theta \propto \mathrm{I}
$
So, the deflection in the galvanometer measures the current I passing through the circuit (ammeter). Shunt resistance is connected in parallel to galvanometer. Therefore, resistance of ammeter can be determined by computing the effective resistance, which is
$
\frac{1}{\mathrm{R}_{e f f}}=\frac{1}{\mathrm{R}_g}+\frac{1}{\mathrm{~S}} \Rightarrow \mathrm{R}_{e f f}=\frac{\mathrm{R}_g \mathrm{~S}}{\mathrm{R}_g+\mathrm{S}}=\mathrm{R}_a
$
Since, the shunt resistance is a very low resistance and the ratio $\frac{S}{R_g}$ is also small. This means, $\mathrm{R}_{\mathrm{g}}$ is also small, i.e., the resistance offered by the ammeter is small. So, when we connect ammeter in series, the ammeter will not change the resistance appreciably and also the current in the circuit. For an ideal ammeter, the resistance must be equal to zero. Hence, the reading in ammeter is always lesser than the actual current in the circuit. Let $\mathrm{I}_{\mathrm{ideal}}$ be current measured from ideal ammeter and $\mathrm{I}_{\text {actual }}$ be the actual current measured in the circuit by the ammeter. Then, the percentage error in measuring a current through an ammeter is
$
\frac{\Delta I}{I} \times 100 \%=\frac{I_{\text {ideal }}-I_{\text {actual }}}{I_{\text {actual }}} \times 100 \%
$

(ii) Galvanometer to a voltmeter: A voltmeter is an instrument used to measure potential difference across any two points in the electrical circuits. It should not draw any current from the circuit otherwise the value of potential difference to be measured will change. Voltmeter must have high resistance and when it is connected in parallel, it will not draw appreciable current so that it will indicate the true potential difference.

A galvanometer is converted into a voltmeter by connecting high resistance $R_h$ in series with galvanometer. The scale is now calibrated in volt and the range of voltmeter depends on the values of the resistance connected in series i.e. the value of resistance is so adjusted that only current $\mathrm{I}_{\mathrm{g}}$ produces full scale deflection i nthe galvanometer.

Let $\mathrm{Rg}$ be the resistance of galvanometer and ' $\mathrm{g}$ be the current with which the galvanometer produces full scale deflection. Since the galvanometer is connected in series with high resistance, the current in the electrical circuit is same as the current passing through the galvanometer.
Potential difference
$
\begin{aligned}
& \mathrm{I}=\mathrm{I}_{\mathrm{g}} \\
& \mathrm{I}=\mathrm{I}_g \Rightarrow \mathrm{I}_g=\frac{\text { Potential difference }}{\text { total resistance }}
\end{aligned}
$
Since the galvanometer and high resistance are connected in series, the total resistance or effective resistance gives the resistance of voltmeter. The voltmeter resistance is $R_V R_g+R_h$
Therefore, $\mathrm{I}_g=\frac{\mathrm{V}}{\mathrm{R}_g+\mathrm{R}_v} \Rightarrow \mathrm{R}_h=\frac{\mathrm{V}}{\mathrm{I}_g}-\mathrm{R}_g$
Note that $\mathrm{I}_{\mathrm{g}} \propto \mathrm{V}$
The deflection in the galvanometer is proportional to current $\mathrm{I}$. But current is proportional to the potential difference. Hence the deflection in the ga'anometer is proportional to potential difference. Since the resistance of voltmeter is very large, a voltmeter connected in an electrical circuit will draw least current in the circuit. An ideal voltmeter is one which has infinite resistance.

Question 12.
Calculate the magnetic field inside and outside of the long solenoid using Ampere's circuital law.
Answer:
Magnetic field due to a long current carrying solenoid:
Consider a solenoid of length $\mathrm{L}$ having $\mathrm{N}$ turns. The diametre of the solenoid is assumed to be much smaller when compared to its length and the coil is wound very closely.

In order to calculate the magnetic field at any point inside the solenoid, we use Ampere's circuital law. Consider a rectangular loop abed. Then from Ampere's circuital law,

$
\begin{aligned}
& \oint_{\mathrm{C}} \overrightarrow{\mathrm{B}} \cdot d \vec{l}=\mu_0 \mathrm{I}_{\text {enclosed }} \\
& =\mu_0 \mathrm{x} \text { (total current enclosed by Amperian loop) } \\
& \text { The left hand side of the equation is } \\
& \oint_{\mathrm{C}} \overrightarrow{\mathrm{B}} \cdot d \vec{l}=\int_a^b \overrightarrow{\mathrm{B}} \cdot d \vec{l}+\int_b^c \overrightarrow{\mathrm{B}} \cdot d \vec{l}+\int_c^d \overrightarrow{\mathrm{B}} \cdot d \vec{l}+\int_d^a \overrightarrow{\mathrm{B}} \cdot d \vec{l}
\end{aligned}
$
Since the elemental lengths along bc and da are perpendicular to the magnetic field which is along the axis of the solenoid, the integrals
$
\begin{aligned}
& \int_b^c \overrightarrow{\mathrm{B}} \cdot d \vec{l}=\int_b^c|\overrightarrow{\mathrm{B}}||d \vec{l}| \cos 90^{\circ}=0 \\
& \int_b^a \overrightarrow{\mathrm{B}} \cdot d \vec{l}=0
\end{aligned}
$
Since the magnetic field outside the solenoid is zero, the integral$
\int_c^d \vec{B} \cdot d \vec{l}=0
$
For the path along ab, the integral is
$
\int_a^b \overrightarrow{\mathrm{B}} \cdot d \vec{l}=\mathrm{B} \int_a^b d l \cos 0^{\circ}=\mathrm{B} \int_a^b d l
$
where the length of the loop ab is $h$. But the choice of length of the loop ab is arbitrary. We ean take very large loop such that it is equal to the length of the solenoid $L$. Therefore the integral is
$
\int_a^b \overrightarrow{\mathrm{B}} \cdot d \vec{l}=\mathrm{BL}
$
let $\mathrm{N}$ I be the current passing through the solenoid of $\mathrm{N}$ turns, then
$
\int_a^b \overrightarrow{\mathrm{B}} \cdot d \vec{l}=\mathrm{BL}=\mu_o \mathrm{NI} \Rightarrow \mathrm{B}=\mu_o \frac{\mathrm{NI}}{\mathrm{L}}
$
The number of turns per unit length is given by $\frac{N I}{L}=\mathrm{n}$, then
$
\mathrm{B}=\mu_0 \frac{n L I}{L}=\mu_0 \mathrm{nI}
$
Since $\mathrm{n}$ is a constant for a given solenoid and $\mathrm{p} 0$ is also constant. For a fixed current $\mathrm{I}$, the magnetic field inside the solenoid is also a constant.