Additional Questions - 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

Additional Questions Solved
I. Choose the Correct Answer
Question 1.
Source (s) of a magnetic field is (are)-
(a) an isolated magnetic pole
(b) a static electric charge
(c) a moving electric charge
(d) all of these
Answer:
(c) a moving electric charge
Question 2.
Magnetic field lines-
(a) cannot intersect at all
(b) intersect at infinity
(c) intersect within the magnet
(d) intersect at the neutral points
Answer:
(a) cannot intersect at all
Question 3.
A magnetic needle is kept in a non-uniform magnetic field. It experiences-
(a) a force and a torque
(b) a force but not a torque
(c) a torque but not a force
(d) neither a force nor a torque
Answer:
(a) a force and a torque
Question 4.
The SI unit of pole strength is-
(a) $\mathrm{Am}^2$
(b) $\mathrm{Am}^{-1}$
(c) $\mathrm{Am}^{-2}$
(d) $\mathrm{Am}$
Answer:
(d) Am
Question 5.
Earth's magnetic field always has a horizontal component except at-
(a) equator

(b) magnetic pole
(c) a latitude of $60^{\circ}$
(d) a latitude of $50^{\circ}$
Answer:
(b) magnetic pole
Question 6.
The angle of dip at the magnetic equator is-
(a) $0^{\circ}$
(b) $30^{\circ}$
(c) $60^{\circ}$
(d) $90^{\circ}$
Answer:
(a) $0^{\circ}$
Question 7.
At a certain place the horizontal component of earth's magnetic field is $\sqrt{3}$ times vertical component. The angle of dip at that place is-
(a) $75^{\circ}$
(b) $60^{\circ}$
(c) $45^{\circ}$
(d) $30^{\circ}$
Answer:
(d) $30^{\circ}$
Hint:
$
\tan \delta=\frac{B_V}{B_H}=\frac{1}{\sqrt{ } 3} \Rightarrow \delta=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=30^{\circ}
$
Question 8.
At magnetic poles the angle of dip is-
(a) $45^{\circ}$
(b) $30^{\circ}$
(c) $0^{\circ}$
(d) $90^{\circ}$
Answer:
(d) $90^{\circ}$
Question 9.
The horizontal component of earth's magnetic field at a place is $3.6 \times 10^{-5} \mathrm{~T}$. If the angle of dip at this place is $60^{\circ}$. the vertical components of earth's field at this place is-
(a) $1.2 \times 10^{-5} \mathrm{~T}$
(b) $2.4 \times 10^{-5} \mathrm{~T}$
(c) $4 \times 10^{-5} \mathrm{~T}$
(d) $6.2 \times 10^{-5} \mathrm{~T}$
Answer:

(d) $6.2 \times 10^{-5} \mathrm{~T}$
Hint:
$
\begin{aligned}
& \mathrm{B}_{\mathrm{v}}=\mathrm{B}_{\mathrm{H}} \tan \delta=3.6 \times 10^{-5} \times \tan 60^{\circ} \\
& \mathrm{B}_{\mathrm{v}}=6.2 \times 10^{-5} \mathrm{~T}
\end{aligned}
$
Question 10.
A bar magnet of magnetic moment $M$ is cut into two parts of equal length. The magnetic moment of either part is-
(a) $\mathrm{M}$
(b) $2 \mathrm{M}$
(c) $\frac{M}{2}$
(d) Zero
Answer:
(c) $\frac{M}{2}$
Question 11.
A magnetic needle suspended by a silk thread is vibrating in the earths magnetic field, If the temperature of the needle is increased by $500^{\circ} \mathrm{C}$, then-
(a) time period decreases
(b) time period remains unchanged
(c) time period increases
(d) the needle stops vibrating
Answer:
(c) time period increases
Hint:
Magnet moment decreases with temperature. Therefore the time period will increase.
Question 12.
Demagnetisation of a magnet can be done by-
(a) rough handling
(b) magnetising in the opposite direction
(c) heating
(d) all the above
Answer:
(d) all the above

Question 13.
All the magnetic materials lose their magnetic properties when-
(a) dipped in water
(b) dipped in oil
(c) brought near a piece of iron
(d) strongly heated
Answer:
(d) strongly heated
Question 14.
The relative permeability of a paramagnetic material is-
(a) greater than unity
(b) less than unity
(c) equal to unity
(d) negative
Answer:
(a) greater than unity
Question 15.
The relative permeability of a diamagnetic material is-
(a) greater than unity
(b) less than unity
(c) equal to unity
(d) negative
Answer:
(b) less than unity
Question 16.
Which of the following is most suitable for the core of an electromagnet?
(a) air
(b) soft iron
(c) steel
(d) Co-Ni alloy
Answer:
(b) soft iron

Question 17.
Soft Iron is used in many parts of electrical machines for-
(a) low hysteresis loss and low permeability
(b) low hysteresis less and high permeability
(c) high hysteresis loss and low permeability
(d) high hysteresis loss and high permeability
Answer:
(b) low hysteresis loss and high permeability
Question 18.
When a diamagnetic substance is brought near the north or the south pole of a bar magnet, it is-
(a) attracted by the poles
(b)repel led by the poles
(c) attracted by the north pole and repelled by the south pole
(d) attracted by the south pole and repelled by the north pole
Answer:
(b) repelled by the poles
Question 19.
At Curie point, a ferromagnetic material becomes-
(a) non magnetic
(b) diamagnetic
(c) paramagnetic
(d) antiferromagnetic
Answer:
(c) paramagnetic
Question 20.
Magnetic permeability is maximum for-
(a) diamagnetic substances
(b) paramagnetic substances
(c) ferromagnetic substances
(d) all of these
Answer:
(c) ferromagnetic substances
Question 21.
The material of a permanent magnet has-
(a) high retentivity, low coercivity
(b) low retentivity, low coercivity
(c) low retentivity, low coercivity

(d) high retentivity, high eoercivity
Answer:
(d) high retentivity, high cuercivity
Question 22.
Which one of the following is not made of soft iron?
(a) electromagnet
(b) core of transformer
(c) core of dynamo
(d) magnet of loudspeaker
Answer:
(d) magnet of loudspeaker
Question 23.
A dip circle $\mathrm{k}$ at right angles to the magnetic meridian. The apparent dip is-
(a) $0^{\circ}$
(b) $30^{\circ}$
(c) 600
(d) $90^{\circ}$
Answer:
(d) $90^{\circ}$
Question 24.
A magnetic needle is placed in a uniform magnetic field. It experience-
(a) a force and a torque
(b) a force but not a torque
(c) a torque but not a force
(d) neither a force nor a torque
Answer:
(c) a torque but not a force
Question 25.
Relative permeability of iron is 5500 . Its magnetic susceptibility is-
(a) 5501
(b) $5500 \times 10^{-7}$
(c) $5500 \times 10^7$
(d) 5499
Answer:
(d) 5499
Hint:
$
\begin{aligned}
& \mu_{\mathrm{r}}=1+\mathrm{x} \\
& \Rightarrow \mathrm{x}=\mu_{\mathrm{r}}-1=5500-1=5499
\end{aligned}
$

Question 26.
The inherent property of all matter is / are-
(a) paramagnetism
(b) diamagnetism
(c) ferromagnetism
(d) all the above
Answer:
(b) diamagnetism
Question 27.
A magnetic needle suspended freely-
(a) orients itself in a definite direction
(b) remains in any direction
(c) become vertical with $\mathrm{N}$-pole up
(d) become vertical with N-pole down
Answer:
(a) orients itself in a definite direction
Question 28.
The earth's magnetic field at a given point is $0.5 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^2$. This field is to be annulled by magnetic induction at the center of circular conducting loop of radius $5 \mathrm{~cm}$. The current required to be flown in the loop is nearly-
(a) $0.2 \mathrm{~A}$
(b) $0.4 \mathrm{~A}$
(c) $4 \mathrm{~A}$
(d) $40 \mathrm{~A}$
Answer:
(b) $0.4 \mathrm{~A}$
Hint:
$
\mathrm{B}=\frac{\mu_0 I}{2 r} \Rightarrow \mathrm{I}=\frac{2 r B}{\mu_0}=\frac{2 \times 0.05 \times 0.5 \times 10^{-5}}{4 \pi \times 10^{-7}}=04 \mathrm{~A}
$
Question 29.
A frog can be leviated in a magnetic field produced by a current in a vertical solenoid placed below the frog. This is possible because the body of the frog is-
(a) paramagnetic
(b) diamagnetic
(c) ferromagnetic
(d) antiferromagnetic
Answer:

(a) paramagnetic

Question 30.
The magnetic moment of a current carrying circular coil of ladius $r$ varies as-
(a) $\frac{1}{r^2}$
(b) $\frac{1}{r}$
(c) $r$
(d) $r^2$
Answer:
(d) $r^2$
Hint:
Magnetic moment, $M=1 A=I\left(\pi r^2\right)=M \alpha r^2$
Question 31.
For a paramagnetic material, the dependence of the magnetic susceptibility $\%$ on the absolute temperature $T$ is given as-
(a) $x \propto r$
(b) $\mathrm{x} \propto \frac{1}{T^2}$
(c) $\mathrm{x} \propto \frac{\frac{T}{T}}{\frac{1}{2}}$
$(\mathrm{d}) \mathrm{x} \propto \mathrm{T}^2$
Answer:
(c) $\mathrm{x} \propto \frac{1}{T}$
Question 32.
A charged particle (charge q) is moving in a circle of radius $\mathrm{R}$ with uniform speed V. The associated magnetic moment is given by-
(a) $\mathrm{qVR}^2$
(b) $\frac{q V R^2}{2}$
(c) $q$ VR
(d) $\frac{q V R}{2}$
Answer:
(d) $\frac{q V R}{2}$
Hint:
Magnetic moment, $\mathrm{M}=\mathrm{IA}=\frac{q}{T}\left(\pi \mathrm{R}^2\right)$
$\mathrm{M}=\frac{q v}{2 \pi R}\left(\pi \mathrm{R}^2\right)=\frac{q v R}{2}$

Question 33.
Nickel shows ferromagnetic property at room temperature. If the temperature is increased . beyond Curie temperature, then it will show-
(a) antiferromagnetism
(b) no magnetic property
(c) diamagnetism
(d) para magnetism
Answer:
(d) paramagnetism
Question 34.
Aproton enters a magnetic field of flux density $1.5 \mathrm{~Wb} / \mathrm{m}^2$ with a speed of $2 \times 10^7 \mathrm{~m} / \mathrm{s}$ at angle of $30^{\circ}$ with the field. The force on the proton will be-
(a) $0.24 \times 10^{-12} \mathrm{~N}$
(b) $2.4 \times 10^{-12} \mathrm{~N}$
(c) $24 \times 10^{-12} \mathrm{~N}$
(d) $0.024 \times 10^{-12} \mathrm{~N}$
Answer:
(b) $2.4 \times 10^{-12} \mathrm{~N}$
Hint:
$\mathrm{F}=$ Bqv $\sin \theta=1.5 \times 1.6 \times 10^{-19} \times 2 \times 10^7 \times \sin 30^{\circ}=2.4 \times 10^{-12} \mathrm{~N}$
Question 35.
A moving charge produces-
(a) an electric field only
(b)a magnetic field only
(c) both electric and magnetic fields
(d) neither an electric nor a magnetic field
Answer:
(c) both electric and magnetic fields
Question 36.
A straight conductor carrying a current $I$, is split into a circular loop of radius $r$ as shown in the figure. The magnetic field at the centre $\mathrm{O}$ of the circle, in tesla is-

(a) $\frac{\mu_0 I}{2 r}$
(b) $\frac{\mu_0 I}{2 \pi r}$
(c) $\frac{\mu_0 I}{\pi r}$
(d) zero
Answer:
(d) zero
Hint:
Field due to the upper and lower semicircles will cancel out.
Question 37.
In a moving coil galvanometer the current ' $i$ ' is related to the deflection $\theta$ as-
(a) $i \alpha \theta$
(b) i $\alpha \tan \theta$
(c) $i \alpha \theta^2$
(d) $\mathrm{i} \alpha \sqrt{ } \theta$
Answer:
(a) $i \alpha \theta$
Question 38 .
The magnetic field due to a current carrying circular coil on the axis, at a large distance $\mathrm{r}$ from the centre of the coil, varies approximately as-
(a) $\frac{1}{r}$
(b) $\frac{1}{r^{\frac{3}{2}}}$
(c) $\frac{1}{r^3}$
(d) $\frac{1}{r^2}$
Answer:
(c) $\frac{1}{r^3}$
Question 39.
A moving charge is subjected to an external magnetic field. The change in the kinetic energy of the particle-
(a) increases with the increase in the field strength
(b) decreases with the increase in the field strength
(c) is always zero
(d) depends upon whether the field is uniform or non-uniform
Answer:
(c) is always zero
Question 40.
Lorentz force is given by-
(a) $q(E+V \times B)$
(b) $q(E-V \times B)$
(c) $q(E+V . B)$
(d) $\mathrm{q}(\mathrm{E} \times \mathrm{B} \times \mathrm{V})$
Answer:
(a) $q(E+V \times B)$
Question 41.
A circular loop has radius $R$ and a current I flows through it. Another circular loop has radius $2 \mathrm{R}$ and a current 21 flows through it. Ratio of the magnetic fields at their centres is-
(a) $\frac{1}{4}$
(b) 1
(c) 2
(d) 4
Answer:
(b) 1
Hint:
$\mathrm{B}_1=\frac{\mu_0 I}{2 r}$ and $\mathrm{B}_2=\frac{\mu(2 I)}{2(2 R)} ; \frac{B_1}{B_2}=1$
Question 42.
The magnetic field inside a solenoid is-
(a) directly proportional to current
(b) inversely proportional to current
(c) directly proportional to its length
(d) inversely proportional to the total number of turns
Answer:
(a) directly proportional to current
Question 43.
A circular loop of area $0.01 \mathrm{~m}^2$ and carrying a current of $10 \mathrm{~A}$ is placed parallel to a magnetic field of intensity $0.1 \mathrm{~T}$. The torque acting on the loop, in $\mathrm{Nm}$ is-
(a) 1.1
(b) 0.8
(c) 0.001

(d) 0.01
Answer:
(d) 0.01
Hint:
$
\tau=\mathrm{BIA}=0.1 \times 10 \times 0.01=0.01 \mathrm{Nm}
$
Question 44.
In a current carrying long solenoid the field produced does not depend upon-
(a) number of turns per unit length
(b) current flowing
(c) radius of the solenoid
(d) all of the above
Answer:
(c) radius of the solenoid
Question 45.
When a charged particle enters a uniform magnetic field its kinetic energy-
(a) remains constant
(b) increases
(c) decreases
(d) becomes zero
Answer:
(a) remains constant
II Fill in the blanks
Question 1.
At Curie point, a ferromagnetic material becomes ..................
Answer:
paramagnetic
Question 2.
Electromagnets are made of soft iron because soft iron has ...................
Answer:
high susceptibility and low retentivity
Question 3.
The word magnetism is derived from iron ore ..............
Answer:
haematetite

Question 4.
The chemical formula of magneties is ................
Answer:
$\mathrm{Fe}_3 \mathrm{O}_4$
Question 5.
The iron ore magnetite was found in the island of ...................
Answer:
Magnesia
Question 6.
suggested that earth behaves as a giant bar magnet ...........
Answer:
Gilbert

Question 7.
The field at the surface of the earth is approximately equal to ...................
Answer:
$
10^{-4} \mathrm{~T}
$
Question 8.
The natural magnets have ...........................
Answer:
irregular shape and they are weak
Question 9.
Pieces of iron or steel that acquires magnetic properties when it is rubbed with a magnet are called ..................
Answer:
artificial magnet
Question 10 .
Artificial magnet in the form of a rectangular or cylinderical bar is called ..................
Answer:
bar magnet
Question 11.
In magnet, the magnetic attraction is maximum at the ...................... of the magnet.
Answer:
poles
Question 12.
The unit of pole strength is ...............
Answer:
ampere meter
Question 13.
The unit of magnetic flux density is.. ....................
Answer:
Weber metre $^2$ (or) tesla
Question 14 .
The value of po is equal to ....................
Answer:
$4 \pi \times 10^{-7} \mathrm{Hm}^{-1}$
Question 15.
A unit pole experiences a force of .......................
Answer:
$
{ }_{10^{-7}} \mathrm{~N}
$

Question 16.
At neutral points the resultant magnetic field due to the magnet and earth is .............
Answer:
magnetic field of the earth
Question 17.
The circular scale of deflection magneto meter is divided into .............. 

Answer:
4
Question 18 .
Each quadrant of the circular scale of deflection magnetometer is graduated from ...............
Answer:
$0-90^{\circ}$
Question 19.
The sensitivity of the deflection magneto meter is more at ..................
Answer:
$45^{\circ}$
Question 20.
The magnetic field used to magnetise a material is called the ..........................
Answer:
magnetic intensity
Question 21.
The unit of magnetising field or magnetic intensity is ..................
Answer:
ampere metre 1
Question 22.
The pole strength per unit area of the cross-section of the material is termed as ..............
Answer:
intensity of magnetisation
Question 23.
The magnetic moment per unit volume of the material is termed as ..................
Answer:
intensity of magnetisation

Question 24.
In diamagnetic materials the net magnetic moment of atoms is ..................
Answer:
zero
Question 25.
The susceptibility of diamagnetic materials has a ................... value.
Answer:
low negative
Question 26.
The susceptibility $\left(\mathrm{x}_{\mathrm{m}}\right)$ of bismuth is ...............
Answer:
0.00017
Question 27.
The relative permeability of diamagnetic material is ...................
Answer:
less than one
Question 28.
Ferromagnetic substances have ..................... magnetic moment.
Answer:
spontaneous net
Question 29.
As the temperature increase, the value of susceptibility of the ferromagnetic substance ........................
Answer:
decrease
Question 30.
The phenomenon of lagging of magnetic induction behind the magnetising field is called ............
Answer:
hysteresis
Question 31 .
The direction of the magnetic field in a current carrying conductor is given by.............
Answer:
Maxwell's right cork screw rule

Question 32.
The relative permeability $\left(\mu_{\mathrm{r}}\right)$ for air is ...............
Answer:
one
Question 33.
The instrument used for measuring current is ...........
Answer:
Tangent galvanometer
Question 34.
Tangent galvanometer works on the principle of ..............
Answer:
Tangent Law
Question 35.
The tangent law is applied in .................
Answer:
Tangent
III Match the following
Question 1.

Answer:
(i) $\rightarrow$ (c)
(ii) $\rightarrow$ (d)
(iii) $\rightarrow$ (a)
(iv) $\rightarrow$ (b)
Question 2.

Answer:
(i) $\rightarrow$ (c)
(ii) $\rightarrow$ (d)
(iii) $\rightarrow$ (b)
(iv) $\rightarrow$ (a)
Question 3.

Answer:
(i) $\rightarrow$ (c)
(ii) $\rightarrow$ (a)
(iii) $\rightarrow$ (d)
(iv) $\rightarrow$ (b)
Question 4.

Answer:

(i) $\rightarrow$ (d)
(ii) $\rightarrow$ (c)
(in) $\rightarrow$ (a)
(iv) $\rightarrow$ (b)
Question 5 .

Answer:
(i) $\rightarrow$ (d)
(ii) $\rightarrow$ (a)
(iii) $\rightarrow$ (b)
(iv) $\rightarrow$ (c)
IV Assertion and Reason Questions
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
(c) If assertion is true but reason is false.
(d) If assertion is false but reason is true.
(e) If both assertion and reason are false.
Question 1.
Assertion: The poles of magnet cannot be separated by breaking into two pieces.
Reason: The magnetic moment will be reduced to half when a magnet is broken into two equal pieces. Answer:
(b) If both assertion and reason are true but reason is not the correct explanation of assertion. Explanation: As we know every atom of a magnet acts as a dipole, so poles cannot be separated. When magnet is broken into two equal pieces, magnetic moment of each part will be half of the original magnet.
Question 2.
Assertion: When radius of circular loop carrying current is doubled, its magnetic moment becomes four times.
Reason: Magnetic moment depends on area of the loop.
Answer:
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
Explanation: Magnetic moment $\mathrm{M}=\mathrm{IA}=\mathrm{I}\left(\pi^2\right)$
New magnetic moment $\mathrm{Ml}=\mathrm{I} \pi(2 \mathrm{r})^2=4 \pi \mathrm{Ir}^2=4 \mathrm{M}$

Question 3.
Assertion: The ferromagnetic substance do not obey Curie's Law.
Reason: At Curie point a ferromagnetic substance start behaving as a paramagnetic substance.
Answer:
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
Explanation: The susceptibility of ferromagnetic substance decreases with the rise of temperature in a complicated manner. After Curie point the susceptibility ferromagnetic substance varies inversely with its absolute temperature.
Question 4.
Assertion: Soft iron is used as transformer core.
Reason: Soft iron has narrow hysteresis loop.
Answer:
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
Explanation: For high efficiency of transformer, the energy loss will be lesser if the hysteresis loop is lesser area, i.e narrow.
Question 5.
Assertion: Cyclotron does not accelerate electron.
Reason: Mass of the electron is very small
Answer:
(a) If both assertion and reason are true and reason is the correct explanation of assertion.
Explanation: Cyclotron is suitable for accelerating heavy particles like protons, x-particles. etc. and not for electrons because of low mass.
Question 6.
Assertion: Cyclotron is a device which is used to accelerate the positive ion.
Reason: Cyclotron frequency depends upon the velocity.
Answer:
(c) If assertion is true but reason is false.
Explanation: Cyclotron is utilised to accelerate the positive ion. And cyclotron frequency is given by $v=$ $\frac{B e}{2 \pi m}$. It means cyclotron frequency doesn't depends upon velocity.

2-mark Questions
Question 1.
Define magnetic declination (D).
Answer:
The angle between magnetic meridian at a point and geographical meridian is called the declination or magnetic declination (D).
Question 2.
Define magnetic inclination (I).
Answer:
The angle subtended by the Earth's total magnetic field B with the horizontal direction in the magnetic meridian is called dip or magnetic inclination (I) at the point.
Question 3.
Define magnetic field.
Answer:
The magnetic field $\vec{B}$ at a point is defined as a force experienced by the bar magnet of unit pole strength. $\mathrm{B}=\frac{1}{q_m} \vec{F}$ Its unit is $\mathrm{N} \mathrm{A}^{-1} \mathrm{~m}^{-1}$
Question 4.
Define magnetic flux density.
Answer:
The magnetic flux density is defined as the number of magnetic field lines crossing unit area kept normal to the direction of line of force. Its unit is $\mathrm{Wb} \mathrm{m}^{-2}$ or tesla.
Question 5.
State 'Tangent Law'.
Answer:
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.
Question 6.
Define in terms of magnetising field.
Answer:
The magnetic field which is used to magnetize a sample or specimen is called the magnetising field. Magnetising field is a vector quantity and it denoted by $\vec{H}$ and its unit is $\mathrm{A} \mathrm{m}^{-1}$.

Question 7.
State Right hand thumb rule.
Answer:
If we curl the fingers of right hand in the direction of current in the loop, then the stretched thumb gives the direction of the mag netic moment associated with the loop.
Question 8.
State 'One ampere'.
Answer:
One ampere is defined as that current when it is passed through each of the two infinitely long parallel straight conductors kept at a distance of one meter apart in vacuum causes each conductor to experience a force of $2 \times 10^{-7}$ newton per meter length of conductor.
3-Mark Questions
Question 1.
Explain Curie's Law of magnetism.
Answer:
When temperature is increased, thermal vibration will upset the alignment of magnetic dipole moments. Therefore, the magnetic susceptibility decreases with increase in temperature. In many cases, the susceptibility of the materials is This relation is called Curie's law.
$\mathrm{x}_{\mathrm{m}} \propto \frac{1}{T}$ or $\mathrm{x}_{\mathrm{m}}=\frac{C}{T}$
This relation is called Curie's law.
Question 2.
Write down the concept of Maxwell's right hand cork screw rule.
Answer:
This rule is used to determine the direction of the magnetic field. If we rotate a right-handed screw using a screw driver, then the direction of current is same as the direction in which screw advances and the direction of rotation of the screw gives the direction of the magnetic field.
Question 3.
Define in terms of Voltage Sensitivity of a galvanometer.
Answer:
It is defined as the deflection produced per unit voltage applied across it.

$
\begin{aligned}
& \mathrm{V}_{\mathrm{s}}=\frac{\theta}{\mathrm{V}} \\
& \mathrm{V}_{\mathrm{s}}=\frac{\theta}{\mathrm{IR}_g}=\frac{\mathrm{NAB}}{\mathrm{KR}_g} \Rightarrow \mathrm{V}_s=\frac{1}{\mathrm{GR}_g}=\frac{\mathrm{I}_s}{\mathrm{R}_g} \\
&
\end{aligned}
$
Question 4.
Define in terms of Current Sensitivity of a galvanometer.
Answer:
It is defined as the deflection produced per unit current flowing through it.
$
\mathrm{I}_{\mathrm{S}}=\frac{\theta}{I}=\frac{N A B}{K} \Rightarrow \mathrm{I}_{\mathrm{S}}=\frac{1}{G}
$
5-Marks Questions

Question 1.
Derive an expression for potential energy of a bar magnet in a uniform magnetic field.
Answer:
When a bar magnet (magnetic dipole) of dipole moment $\vec{B}_{\mathrm{m}}$ is held at an angle 0 with the direction of a uniform magnetic field $B$, the magnitude of the torque acting on the dipole is

$\left|\vec{\tau}_B\right|=\left|\vec{P}_B\right||\vec{B}| \sin \theta \ldots \ldots$ (1)
If the dipole is rotated through a very small angular displacement $d \theta$ against the torque $\tau_B$ at constant angular velocity, then the work done by external torque $\left(\vec{\tau}_{e x t}\right)$ for this small angular displacement is given by $\mathrm{dW}=\left(\vec{\tau}_{e x t}\right) \mathrm{d} \theta$..... (2)
The bar magnet has to be moved at constant angular velocity, which implies that $\left|\vec{\tau}_B\right|=\left|\vec{\tau}_{e x t}\right|$ $\mathrm{dW}=\mathrm{P}_{\mathrm{m}} \mathrm{B} \sin \theta \mathrm{d} \theta$
Total work done in rotating the dipole from $\theta$ ' to $\theta$ is e e
$
\begin{aligned}
& \mathrm{W}=\int_{\theta^{\prime}}^\theta \tau d \theta=\int_{\theta^{\prime}}^\theta \mathrm{P}_m \mathrm{~B} \sin \theta d \theta=\mathrm{P}_m \mathrm{~B}[-\cos \theta d \theta]_{\theta^{\prime}}^\theta . \\
& \mathrm{W}=-\mathrm{P}_{\mathrm{m}} \mathrm{B}\left(\cos \theta-\cos \theta^{\prime}\right) \ldots . . \text { (3) }
\end{aligned}
$
This work done is stored as potential energy in bar magnet at an angle $\theta$ when it is rotated from $\theta^{\prime}$ to $\theta$ and it can be written as $\mathrm{U}=-\mathrm{P}_{\mathrm{m}} \mathrm{B}\left(\cos \theta-\cos \theta^{\prime}\right)$
In fact, the equation (4) gives the difference in potential energy between the angular positions $\theta^{\prime}$ and $\theta$. We can choose the reference point $\theta^{\prime}=90^{\circ}$, so that second term in the equation becomes zero and the equation 4 can be written as $\mathrm{U}=-\mathrm{P}_{\mathrm{m}} \mathrm{B}(\cos \theta)$
$
\mathrm{U}=-\mathrm{P}_{\mathrm{m}} \cdot \vec{B}
$
Case 1:
(i) If $\theta=0^{\circ}$, then
$\mathrm{U}=-\mathrm{P}_{\mathrm{m}} \mathrm{B}\left(\cos 0^{\circ}\right)=\mathrm{P}_{\mathrm{m}} \mathrm{B}$
(ii) If $\theta=180^{\circ}$, then

$\mathrm{U}=-\mathrm{p}_{\mathrm{m}} \mathrm{B}\left(\cos 180^{\circ}\right)=\mathrm{p}_{\mathrm{m}} \mathrm{B}$
The potential energy stored in a bar magnet in a uniform magnetic field is given by
Question 2.
Write down the application of Hysteresis loop.
Answer:
Applications of hysteresis loop:
The significance of hysteresis loop is that it provides information such as retentivity, coercivity, permeability, susceptibility and energy loss during one cycle of magnetisation for each ferromagnetic material. Therefore, the study of hysteresis loop will help us in selecting proper and suitable material for a given purpose. Some examples:
(i) Permanent magnets:
The materials with high retentivity, high coercivity and high permeability are suitable for making permanent magnets.
Examples:
Steel and Alnico
(ii) Electromagnets:
The materials with high initial permeability, low retentivity, low coercivity and thin hysteresis loop with smaller area are preferred to make electromagnets.
Examples:
Soft iron and Mumetal (Nickel Iron alloy).
(iii) Core of the transformer:
The materials with high initial permeability, large magnetic induction and thin hysteresis loop with smaller area are needed to design transformer cores.
Examples:
Soft iron
Question 3.
Write down the difference between soft and hard ferromagnetic materials.
Answer:

Question 4.
State and explain the Biot - Savart Law.
Answer:
Biot and Savart experimentally observed that the magnitude of magnetic field $\mathrm{d} \vec{B}$ at a point $\mathrm{P}$ at a distance $\mathrm{r}$ 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 $\overrightarrow{d l}$ and $\hat{r}$.
(iv) inversely as the square of the distance between the point $\mathrm{P}$ and length element $\overrightarrow{d l}$. This is expressed as

$
\begin{aligned}
& \mathrm{dB} \propto \frac{I d l}{r^2} \sin \theta \\
& \mathrm{dB}=\mathrm{k} \frac{I d l}{r^2} \sin \theta
\end{aligned}
$
Where $\mathrm{K}=\frac{\mu_0}{4 \pi}$ in SI units and $\mathrm{k}=1$ in CGS units.
In vector notation,
$
\mathrm{d} \vec{B}=\frac{\mu_0}{4 \pi} \frac{\mathrm{I} \vec{l} \times \dot{r}}{r^2}
$
Here vector $\mathrm{d} \vec{B}$ is perpendicular to both I $\vec{d} l$ (pointing current carrying conductor the direction of current flow) and the unit vector and $\hat{r}$ directed from $\vec{d} l$ toward point $P$ The equation 1 is used to compute the magnetic field only due to a small elemental length $\vec{d} l$ of the conductor. The net magnetic field at $P$ due to the conductor is obtained from principle of superposition by considering the contribution from all current elements I $\vec{d} l$. Hence integrating equation (1), we get
$
\vec{B}=\int \mathrm{d} \vec{B}=\frac{\mu_0 I}{4 \pi} \int \frac{\vec{d} \times \quad \hat{r}}{r^2}
$
where the integral is taken over the entire current distribution.
Case:
1. If the pont $P$ lies on the ckonductor, then $\theta=0^{\circ}$. Therefore, $\mathrm{d} \vec{B}$ is zero.
2. If the point lies perpendicular to the conductor, then $\theta=90^{\circ}$. Therefore, $\mathrm{d} \vec{B}$ is maximum and is given by $\mathrm{d} \vec{B}=\frac{I d l}{r 2} \hat{n}$
where $\hat{n}$ is the unit vector perpendicular to both I $\vec{d} l$ and
Question 5.
Write down the difference between Coulomb's Law and Biot-Savart Law
Answer:

Question 6.
Obtain an expression for the magnetic dipole moment of a revolving electron.
Answer:
Magnetic dipole moment of revolving electron:
Suppose an electron undergoes circular motion around the nucleus. The circulating electron in a loop is like current in a circular loop (since flow of charge is current). The magnetic dipole moment due to current carrying circular loop is

$
\vec{\mu}_{\mathrm{L}}=\vec{A}
$
In magnitude, $\mu=\mathrm{I} \mathrm{A}$ If $T$ is the time period of an electron, the current due to circular motion of th electron is $\mathrm{I}=\frac{-e}{T}$
where - $e$ is the charge of an electron. If $R$ is the radius of the circular orbit and $v$ is the velocity of the electron in the circular orbit, then
$
\mathrm{T}=\frac{2 \pi R}{v}
$
Using equation (2) and equation (3) in equation (1), we get
$
\mu_{\mathrm{L}}=\frac{\frac{e}{2 \pi \mathrm{R}}}{v} \pi \mathrm{R}^2=\frac{e v R}{2}
$
Where $\mathrm{A}=\pi \mathrm{R}^2$ is the area of the circular loop. By definetion, angular moment of the electron about $\mathrm{O}$ is $\vec{L}=\vec{R} \times \vec{P}$
In magnitude, $\mathrm{L}=\mathrm{PR}=\operatorname{mvR}$....... (5)
Using equation (4) and equation (5), we get

$
\frac{\mu_L}{L}=-\frac{e v R}{\frac{2}{m v R}}=\frac{e}{2 m} \vec{L}
$
The negative sign indicates that the magnetic moment and angular momentum are in opposite direction. In magnitude,
$
\begin{aligned}
& \frac{\mu_L}{L}=\frac{e}{2 m}=\frac{1.60 \times 10^{-19}}{2 \times 9.11 \times 10^{-31}}=0.0878 \times 10^{12} \\
& \frac{\mu_L}{L}=8.78 \times 10^{10} \mathrm{C} \mathrm{kg}^{-1}=\text { content }
\end{aligned}
$
The ration $\frac{\mu_L}{L}$ is a constant and also known as gyro-magnetic ration $\frac{\mu_L}{L}$. It must be noted that the gyromagnetic ratio is a constant of proportionality which connects angular momentum of the electron and the magnetic moment of the electron. According to Neil's Bohr quantization rule, the angular momentum of an electron moving in a stationary orbit is quantized, which means,
$
1=\mathrm{nh}=\mathrm{n} \frac{h}{2 \pi}
$
where, $\mathrm{h}$ is the planck's constant $\left(\mathrm{h}=6.63 \times 10^{-34} \mathrm{~J}\right.$ s) and number $\mathrm{n}$ takes natural numbers (i.e., $\mathrm{n}=$ $1,2,3, \ldots$.$) . Hence,$
$
\begin{aligned}
& \mu_{\mathrm{L}}=\frac{e}{2 m} \mathrm{~L}=n \frac{e h}{4 \pi m} \mathrm{Am}^2 \\
& \mu_{\mathrm{L}}=n \frac{\left(1.60 \times 10^{-19}\right) h}{4 \pi m} \mathrm{Am}^2=n \frac{\left(1.60 \times 10^{-19}\right)\left(6.63 \times 10^{-34}\right)}{4 \times 3.14 \times\left(9.11 \times 10^{-31}\right)} \\
& \mu_{\mathrm{L}}=\mathrm{n} \times 9.27 \times 10^{-24} \mathrm{Am}^2
\end{aligned}
$
The minimum magnetic moment can be obtained by subsituting by sustituting $\mathrm{n}=1$,
$
\mu_{\mathrm{L}}=\mathrm{n} \times 9.27 \times 10^{-24} \mathrm{~A} \mathrm{~m}^2=9.27 \times 10^{-24} \mathrm{~J}^{-1}=\left(\mu_{\mathrm{L}}\right)_{\min }=\mu_{\mathrm{B}}
$
Where, $\mu_{\mathrm{B}}=\frac{e h}{4 \pi m}=9.27 \times 10^{-24} \mathrm{~A} \mathrm{~m}^2$
is called Bohr magneton. This is a convenient unit with which one can measure atomic magnetic moments.
Question 7.
Apply Ampere's Circuital Law to find the magnetic field both inside and outside of a toroidal solenoid.

Answer:
A solenoid is bent in such a way its ends are joined together to form a closed ring shape, is called a toroid. The magnetic field has constant magnitude inside the toroid whereas in the interior region (say, at point $P$ ) and exterior region (say, at point Q), the magnetic field is zero.
(a) Open space interior to the toroid:
Let us calculate the magnetic field $B_p$ at point $P$. We construct an Amperian loop 1 of radius $r_1$ around the point P. For simplicity, we take circular loop so that the length of the loop is its circumference.

$
\mathrm{L}_1=2 \pi_1
$
Ampere's circuital law for the loop 1 is $\oint_{\text {loop } 1} \overrightarrow{\mathrm{B}}_{\mathrm{P}} \cdot \vec{d} l=\mu_0 \mathrm{I}_{\text {enclosed }}$
Scince, the loop 1 encloses no current, $\mathrm{I}_{\text {enclosed }}=0$
$
\oint_{\text {loop } 1} \overrightarrow{\mathrm{B}}_{\mathrm{P}} \cdot \vec{d} l=0
$
This is possible only if the magnetic field at point $P$ vanishes i.e.
$
\vec{B}_{\mathrm{p}}=0
$
(b) Open space exterior to the toroid:
Let us calculate the magnetic field $B_Q$ at point $Q$. We construct an Amperian loop 3 of radius $r_3$ around the point $Q$. The length of the loop is $L_3=2 \pi r_3$ Ampere's circuital law for the loop 3 is
$
\oint_{\text {loop } 3} \overrightarrow{\mathrm{B}}_{\mathrm{Q}} \cdot \vec{d} l=\mu_0 \mathrm{I}_{\text {enclosed }}
$
Since, in each turn of the toroid loop, current coming out of the plane of paper is cancelled by the current going into the plane of paper. Thus, $\mathrm{I}_{\text {enclosed }}=0$
$
\oint_{\text {loop } 3} \overrightarrow{\mathrm{B}}_{\mathrm{Q}} \cdot \vec{d} l=0
$
This is possible only if the magnetic field at point $\mathrm{Q}$ vanishes i.e. $\vec{B}_{\mathrm{Q}}=0$
(c) Inside the toroid:
Let us calculate the magnetic field $B_S$ at point $S$ by constructing an Amperian loop 2 of radius $r_2$ around the point $\mathrm{S}$. The length of the loop is $\mathrm{L}_2=2 \pi \mathrm{r}_2$ Ampere's circuital law for the loop 2 is

$\oint_{\text {loop } 2} \overrightarrow{\mathrm{B}}_{\mathrm{s}} \cdot \vec{d} l=\mu_0 \mathrm{I}_{\text {enclosed }}$
Let I be the current passing through the toroid and $\mathrm{N}$ be the number of turns of the toroid, then $\mathrm{I}_{\text {enclosed }}=$ NI and $\oint_{\text {loop 2 }} \overrightarrow{\mathrm{B}}_s \cdot \vec{d} l=\oint_{\text {loop 2 }} \mathrm{B} d l \cos \theta=\mathrm{B} 2 \pi r_2$
$
\begin{aligned}
& \oint_{\text {loop 2 }} \overrightarrow{\mathrm{B}}_s \cdot \vec{d} l=\mu_0 \\
& \mathrm{~B}_s=\mu_0 \frac{\mathrm{NI}}{2 \pi r_2}
\end{aligned}
$
Question 8.
Obtain an expression for force on a moving charge in a magnetic field.
Answer:
When an electric charge q is moving with velocity $\vec{v}$ in the magnetic field $\vec{B}$, it experiences a force, called magnetic force $\vec{F}_{\mathrm{m}}$. After careful experiments, Lorentz deduced the force experienced by a moving charge in the magnetic field $\vec{F}_{\mathrm{m}}$

$
\vec{F}_{\mathrm{m}}=\mathrm{q}(\vec{v} \times \vec{B})
$
In magnitude, $\mathrm{F}_{\mathrm{m}}=\mathrm{qvB} \sin \theta$
The equations (1) and (2) imply
1. $\vec{F}_{\mathrm{m}}$ is directly proportional to the magnetic field $\vec{B}$
2. $\vec{F}_{\mathrm{m}}$ is directly proportional to the velocity $\vec{v}$
3. $\vec{F}_{\mathrm{m}}$ is directly proportional to sine of the angle between the velocity and magnetic field
4. $\vec{F}_{\mathrm{m}}$ is directly proportional to the magnitude of the charge $\mathrm{q}$
5. The direction of $\vec{F}_{\mathrm{m}}$ is always perpendicular to $\vec{v}$ and $\mathrm{g}$ as $\vec{F}_{\mathrm{m}}$ is ti'e cross product of $\vec{v}$ and $\vec{B}$
6. The direction of $\mathrm{jprn}$ is on negative charge is opposite to the direction of $\mathrm{F}$ charge provided other factors are identical.
7. If velocity $\mathrm{v}$ of the charge $\mathrm{q}$ is along magnetic field $\vec{B}$ then, $\vec{F}_{\mathrm{m}}$ is zero.