Numerical problems-1 - 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

Numerical problems
Question 1.
A bar magnet having a magnetic moment $\vec{M}$ is cut into four pieces i.e., first cut in two pieces along the axis of the magnet and each piece is further cut into two pieces. Compute the magnetic moment of each piece.
Solution:
Consider a bar magnet of magnetic moment $\vec{M}$. When a bar magnet first cut in two pieces
along the axis, their magnetic moment is $\frac{\bar{M}}{2}$

Their magnetic moment of each pieces $\frac{\vec{M}}{\stackrel{4}{\rightarrow}}$
Their magnetic moment of each pieces $\vec{M}_{\text {new }} \frac{1}{4} \vec{M}$
Question 2.
A conductor of linear mass density $0.2 \mathrm{~g} \mathrm{~m}^{-1}$ suspended by two flexible wire as shown in figure. Suppose the tension in the supporting wires js zero when it is kept inside the magnetic field of $\mathrm{I}$ T whose direction
is into the page. Compute the current inside the conductor and also the direction of the current. Assume $g$ $=10 \mathrm{~m} \mathrm{~s}^{-2}$.

Solution:
Downward force, $\mathrm{F}=\mathrm{mg}$
Linear mass density, $\frac{m}{l}=0.2 \mathrm{gm}^{-1}$
$
\begin{aligned}
& \frac{m}{l}=0.2 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1} \\
& \mathrm{~m}=\left(0.2 \times 10^{-3} \times 1\right) \mathrm{kg} \mathrm{m}^{-1} \\
& \mathrm{~F}=\left(0.2 \times 10^{-3} \times 1 \times 10\right) \mathrm{N}
\end{aligned}
$
Upward magnetic force acting on the wire
$
\begin{aligned}
& \mathrm{F}=\mathrm{BIl} \\
& 0.2 \times 10^{-3} \times 1 \times 10=1 \times \mathrm{I} \times 1 \\
& \mathrm{I}=2 \times 10^{-3} \\
& \mathrm{I}=2 \mathrm{~m} \mathrm{~A}
\end{aligned}
$
Question 3.
A circular coil with cross-sectional area $0.1 \mathrm{~cm}^2$ is kept in a uniform magnetic field of strength $0.2 \mathrm{~T}$. If the current passing in the coil is $3 \mathrm{~A}$ and plane of the loop is perpendicular to the direction of magnetic field. Calculate
(a) total torque on the coil
(b) total force on the coil
(c) average force on each electron in the coil due to the magnetic field of the free electron density for the material of the wire is $10^{28} \mathrm{~m}^{-3}$
Solution:
Cross sectional area of coil, $\mathrm{A}=0.1 \mathrm{~cm}^2$
$
\mathrm{A}=0.1 \times 10^{-4} \mathrm{~m}^2
$
Uniform magnetic field of strength, $\mathrm{B}=0.2 \mathrm{~T}$
Current passing in the coil, $I=3 \mathrm{~A}$
Angle between the magnetic field and normal to the coil, $\theta=0^{\circ}$
(a) Total torque on the coil,
$
\begin{aligned}
& \tau=\mathrm{ABI} \sin \theta=0.1 \times 10^{-4} \times 0.2 \times 3 \sin 0^{\circ} \sin 0^{\circ}=0 \\
& \tau=0
\end{aligned}
$
(b) Total force on the coil
$
\begin{aligned}
& \mathrm{F}=\mathrm{BIl} \sin \theta=0.2 \times 3 \times 1 \times \sin 0^{\circ} \\
& \mathrm{F}=0
\end{aligned}
$
(c) Average force:
$
\mathrm{F}=\mathrm{qV}_{\mathrm{d}} \mathrm{B}
$

$
\begin{aligned}
& \text { drift velocity, } \mathrm{V}_{\mathrm{d}}=\frac{1}{n e A} \\
& {[\because \mathrm{q}=\mathrm{e}]} \\
& \mathrm{F}=\mathrm{e}\left(\frac{1}{n e A}\right) \mathrm{B} \\
& {\left[\because \mathrm{n}=10^{28} \mathrm{~m}^{-3}\right.} \\
& \frac{I B}{n A}=\frac{3 \times 0.2}{10^{28} \times 0.1 \times 10^{-4}}=6 \times 10^{-24} \\
& \mathrm{~F}_{\mathrm{av}}=0.6 \times 10^{-23} \mathrm{~N}
\end{aligned}
$
Question 4.
A bar magnet is placed in a uniform magnetic field whose strength is $0.8 \mathrm{~T}$. Suppose the bar magnet orient at an angle $30^{\circ}$ with the external field experiences a torque of $0.2 \mathrm{~N} \mathrm{~m}$. Calculate:
(i) the magnetic moment of the magnet
(ii) the work done by an applied force in moving it from most stable configuration to the most unstable configuration and also compute the work done by the applied magnetic field in this case.
Solution:
Uniform magnetic field strength $\mathrm{B}=0.8 \mathrm{~T}$
Bar magnet orient an angle with magnetic field $\theta=30^{\circ}$
Torque $\tau=0.2 \mathrm{Nm}$
(i) Magnetic moment of the magnet,
Torque $\tau=\mathrm{P}_{\mathrm{m}} \mathrm{B} \operatorname{Sin} \theta$
$\therefore$ Magnetic moment, $\mathrm{P}_{\mathrm{m}}=\frac{\tau}{B \sin \theta}=\frac{0.2}{0.8 \times \operatorname{Sin} 30^{\circ}}=\frac{0.2}{0.4}$
$\mathrm{P}_{\mathrm{m}}=0.5 \mathrm{Am}^2$
(ii) Work done by external torque is stored in the magnet as potential energy. $\mathrm{W}=\mathrm{U}=-\mathrm{P}_{\mathrm{m}} \mathrm{B} \operatorname{Sin} \theta$
Here, applied force acting on magnet its moving from most stable $\theta^{\prime}$ to most unstable $\theta$.
$\theta^{\prime}=0^{\circ}$ and $\theta=180^{\circ}$
So, workdone $\mathrm{W}=\mathrm{U}=-\mathrm{P}_{\mathrm{m}} \mathrm{B}\left(\operatorname{Cos} \theta-\operatorname{Cos} \theta^{\prime}\right)$
$
\begin{aligned}
& =-\mathrm{P}_{\mathrm{m}} \mathrm{B}\left(\operatorname{Cos} 180^{\circ}-\operatorname{Cos} 0^{\circ}\right)=-0.5 \times 0.8((-1)-1)=-0.4(-2) \\
& \mathrm{W}=\mathrm{U}=0.8 \\
& \mathrm{~W}=0.8 \mathrm{~J}
\end{aligned}
$
Question 5.
A non - conducting sphere has a mass of $100 \mathrm{~g}$ and radius $20 \mathrm{~cm}$. A flat compact coil of wire with turns 5 is wrapped tightly around it with each turns concentric with the sphere. This sphere is placed on an inclined plane such that plane of coil is parallel to the inclined plane. A uniform magnetic field of $0.5 \mathrm{~T}$ exists in the region in vertically upward direction. Compute the current 1 required to rest the sphere in equilibrium.

Solution:
At equilibrium
$\mathrm{f}_{\mathrm{S}} \mathrm{R}-\mathrm{p}_{\mathrm{m}} \mathrm{B} \sin \theta=0$
$\mathrm{mgR}=\mathrm{NBAI}$
$\mathrm{I}=\frac{m g R}{N B A}=\frac{m g R}{N B \pi R^2}$
$\mathrm{I}=\frac{m g}{\pi R N B}$
Mass of the sphere, $\mathrm{m}=100 \mathrm{~g}=100 \times 10^{-3} \mathrm{~kg}$
Radius of the sphere $\mathrm{R}=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}$
Number of turns $\mathrm{n}=5$
Uniform magnetic field $\mathrm{B}=0.5 \mathrm{~T}$
$\mathrm{I}=\frac{100 \times 10^{-3} \times 10}{\pi \times 20 \times 10^{-2} \times 5 \times 0.5}=\frac{1000 \times 10^{-3}}{\pi \times 50 \times 10^{-2}}=\frac{20 \times 10^{-1}}{\pi}$
$\mathrm{I}=\frac{2}{\pi} \mathrm{A}$.
Question 6.
Calculate the magnetic field at the center of a square loop which carries a current of $1.5 \mathrm{~A}$, length of each loop is $50 \mathrm{~cm}$.
Solution:

Current through the square loop, $\mathrm{I}=1.5 \mathrm{~A}$
Length of each loop, $1=50 \mathrm{~cm} 50 \times 10^{-2} \mathrm{~m}$
According to Biot-Savart Law.
Magnetic field due to a current carrying straight wire
$
\begin{aligned}
& B=\frac{\mu_0 I}{4 \pi a}(\operatorname{Sin} \alpha+\operatorname{Sin} \beta)=\frac{4 \pi \times 10^{-7} \times 1.5}{4 \pi \times\left(\frac{l}{2}\right)}\left(\operatorname{Sin} 45^{\circ}+\operatorname{Sin} 45^{\circ}\right) \\
& =\frac{2 \times 1.5 \times 10^{-7}}{l}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)=\frac{2 \times 1.5 \times 10^{-7}}{50 \times 10^{-2}}\left(\frac{2}{\sqrt{2}}\right) \\
& \mathrm{B}=0.084866 \times 10^{-5} \mathrm{~T} \\
& =4 \times 0.08487 \times 10^{-5}=0.33948 \times 10^{-5} \\
& B^{\prime}=3.4 \times 10^{-6} \mathrm{~T} \\
&
\end{aligned}
$
Question 7.
Show that the magnetic field at any point on the axis of the solenoid having $n$ turns per unit length is $B=$

$\frac{1}{2} \mu_0 \mathrm{nI}\left(\cos \theta_1-\cos \theta_2\right)$
Solution:
A solenoid is a cylindrical coil having number of circular turns. Consider a solenoid having radius $R$ consists of $n$ number'of turns per unit length.
Let ' $P$ ' be the point at a distance ' $x$ ' from the origin of the solenoid. The current carrying element d $x$ at a distance $\mathrm{x}$ from origin and the distance $\mathrm{r}$ from point ' $\mathrm{P}$ '.
$
\mathrm{r}=\sqrt{\mathrm{R}^2+\left(x_0-x\right)^2}
$
The magnetic field due to current carrying circular coil at any axis is
$
\mathrm{dB}=\frac{\mu_0 \mathrm{IR}^2}{2 r^3} \times \mathrm{N}
$
Where $\mathrm{N}=\mathrm{ndx}$, then
$
\begin{aligned}
& \mathrm{dB}=\frac{\mu_0}{2} \frac{n \mathrm{IR}^2 d x}{r^3} \ldots \ldots .(1) \\
& \sin \theta=\frac{R}{r} \\
& \mathrm{r}=\mathrm{R} \times \frac{1}{\sin \theta}=\mathrm{R} \operatorname{cosec} \theta \ldots \\
& \tan \theta=\frac{R}{x_0-x} \\
& \mathrm{x}_0-\mathrm{x}=\mathrm{R} \times \frac{1}{\tan \theta}=\mathrm{R} \cot \theta \\
& \frac{d x}{d \theta}=\mathrm{R} \operatorname{cosec}^2 \theta \\
& \Rightarrow \mathrm{dx}=\mathrm{R} \operatorname{cosec}^2 \theta \mathrm{d} \theta \ldots \ldots .
\end{aligned}
$

From above three equations, wer get
$
\begin{aligned}
& \mathrm{dB}=\frac{\mu_0}{2} \frac{n \mathrm{IR}^2\left(\operatorname{Resc}^2 \theta d \theta\right)}{\mathrm{R}^3 \csc ^3 \theta} \\
& \mathrm{dB}=\frac{\mu_0}{2} \mathrm{n} \mathrm{I} \sin \theta \mathrm{d} \theta
\end{aligned}
$
Now total magnetic field can be obtained by integrating from $\theta_1$ to $\theta_2$, we get
$
\begin{aligned}
& \mathrm{B}=\frac{\mu_0 n \mathrm{I}}{2} \int_{\theta_1}^{\theta_2} \sin \theta d \theta=\frac{\mu_0 n \mathrm{I}}{2}[-\cos \theta]_{\theta 1}^{\theta 2} \\
& \mathrm{~B}=-\frac{\mu_0 n \mathrm{I}}{2}\left[\cos \theta_2-\cos \theta_1\right] \\
& \mathrm{B}=\frac{\mu_0 n \mathrm{I}}{2}\left[\cos \theta_1-\cos \theta_2\right]
\end{aligned}
$
Question 8.
Let $\mathrm{I}_1$ and $\mathrm{I}_2$ be the steady currents passing through a long horizontal wire $\mathrm{XY}$ and $P Q$ respectively. The wire $P Q$ is fixed in horizontal plane and the wire $X Y$ be is allowed to move freely in a vertical plane. Let the wire $X Y$ is in equilibrium at a height $d$ over the parallel wire $P Q$ as shown in figure.

Solution:
If the wire $X Y$ is slightly displaced and released, it executes simple uarmonic motion due to force of
repulsion produced between the current carrying wire.
Acceleration of the wire, $a=-\omega^2 y$
Time period of oscillation of the wire,
$
\mathrm{T}=2 \pi \sqrt{\frac{d}{g}}
$