Magnetism And Matter Class 12 Important Questions: CBSE Physics Chapter 5

Very Short Answer Questions (1 Mark)

1. How does the intensity of magnetization of a paramagnetic material vary with increasing applied magnetic field?

Ans: The intensity of magnetization increases with the increase in applied magnetic field.

2. An iron bar magnet is heated to $${1000}^{0}C$$ and then cooled in a magnetic field free space. Will it retain magnetism?

Ans: Curie temperature of iron is $${770}^{0}C$$. When it is heated to a very high temperature magnetism of iron is lost and does not retain its magnetism further.

3. How will the magnetic field intensity at the centre of a circular wire carrying current change, if the current through the wire is doubled and radius of the coil is halved?

Ans: As $$B={\displaystyle \frac{{\mu }_0}{4\pi }}{\displaystyle \frac{2\pi I}{r}}$$

$$\Rightarrow B^{\prime }={\displaystyle \frac{{\mu }_0}{4\pi }}{\displaystyle \frac{2\pi \left(2I\right)}{r/2}}$$

$$\Rightarrow B^{\prime }=4\left({\displaystyle \frac{{\mu }_0}{4\pi }}{\displaystyle \frac{2\pi I}{r}}\right)$$

$$\Rightarrow B^{\prime }=4B$$

4. Can neutrons be accelerated in a cyclotron? Why?

Ans:  No, neutrons cannot be accelerated in a cyclotron. This is because a neutron is neutral and a cyclotron can accelerate only charged particles.

5. What type of magnetic material is used in making permanent magnets?

Ans: Materials having high coercivity are used in making permanent magnets.

6. Which physical quantity has the unit $$Wb/m^2$$ ? Is it a scalar or a vector quantity?

Ans: Magnetic field has the unit $$Wb/m^2$$. It is a vector quantity.

Short Answer Questions (2 Mark)

1. A bar magnet of magnetic moment M is aligned parallel to the direction of a uniform magnetic field B. What is the work done to turn the magnet, so as the align its magnetic moment:

(i). Opposite to the field direction?

Ans: We know that work done, $$W=MB(\cos {\theta }_1-\cos {\theta }_2)$$

Here, ${\theta }_1={\text{0}}^{\text{0}}\text{ and }{\theta }_{\text{2}}\text{=18}{0}^0$

$$\Rightarrow W=MB(\cos 0^0-\cos {180}^0)$$

$$\Rightarrow W=MB(1-(-1))$$

$$\Rightarrow W=\text{ 2}MB$$

(ii). Normal to the field direction?

Ans: Here, ${\theta }_1={\text{0}}^{\text{0}}\text{ and }{\theta }_{\text{2}}\text{=9}{0}^0$

$$\Rightarrow W=MB(\cos 0^0-\cos {90}^0)$$

$$\Rightarrow W=MB$$

2. An electron in the ground state of hydrogen atom is revolving in anti - clock wise direction in a circular orbit. The atom is placed normal to the electron orbit makes an angle of $${30}^0$$ in the magnetic field. Find the torque experienced by the orbiting electron?

Ans: In the above question it is given that:

Magnetic moment associated with electron $$M={\displaystyle \frac{eh}{4\pi m_e}}$$

$\theta ={30}^0$ and

$\tau =MB\sin \theta$

$$\tau ={\displaystyle \frac{eh}{4\pi m_e}}B\times \sin {30}^0={\displaystyle \frac{eh}{8\pi m_e}}$$, which is the torque.

3. Define angle of dip. Deduce the relation connecting angle of dip and horizontal component of earth’s total magnetic field with the horizontal direction.

Ans: We know that:

${\displaystyle \frac{B_H}{B}}=\cos \delta$ and

${\displaystyle \frac{B_V}{B}}=\sin \delta$

$\Rightarrow {\displaystyle \frac{\sin \delta }{\cos \delta }}={\displaystyle \frac{B_V}{B}}\times {\displaystyle \frac{B}{B_H}}$

$\Rightarrow \tan \delta ={\displaystyle \frac{B_V}{B_H}}$

4. A point change +q is moving with speed perpendicular to the magnetic field B as shown in the figure. What should be the magnitude and direction of the applied electric field so that the net force acting on the charge is zero?

Ans: We know that:

Force on the charge due to magnetic field $$=qVBsin\theta$$

Since $\overrightarrow{B}\mathrm{\perp }$ to the plane of paper,

$$F=qVBsin{90}^0$$

$$F=qVB$$ (along OY)

Force on the charge due to electric field is:

$$F=qE$$

Net force on charge is zero if $$qE=qVB$$

$$E=VB$$ (along YO)

5. The energy of a charged particle moving in a uniform magnetic field does not change. Why?

Ans: The force on a charged particle in a uniform magnetic field always acts in a direction perpendicular to the motion of the charge. As the work done by the magnetic field on the charge is zero, hence energy of the charged particle will not change.

6. In the figure, straight wire AB is fixed; white the loop is free to move under the influence of the electric currents flowing in them. In which direction does the loop begin to move? Justify.

Ans: As the current in AB and arm PQ are in the same direction therefore wire will attract the arm PQ with a force (say $F_1$ ). But repels the arm RS with a force (say $F_2$ ). Since arm PQ is closer to the wire AB, $F_1>F_2$ i.e., the loop will move towards the wire.

7. State two factors by which voltage sensitivity of a moving coil galvanometer can be increased.

Ans: We know that:

$$Voltagesensitivity={\displaystyle \frac{nBA}{kr}}$$

It can be increased by

(1) increasing B using powerful magnets.

(2) decreasing k by using phosphor borne strips.

8. What is the magnetic moment associated with a coil of 1 turn, area of cross- section ${10}^{-4}{m}^2$ carrying a current of 2A?

Ans: We know that:

$$m=NIA$$

$$\Rightarrow m=\text{ 1}\times \text{1}{\text{0}}^4\times 2$$

$$\Rightarrow m=\text{ 2}\times \text{1}{\text{0}}^{4}A{m}^2$$

9. A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetising current of 1.2 A?

Ans: In the above question it is given that:

Mean radius of a Rowland ring, $$r=15cm=0.15m$$

Number of turns on a ferromagnetic core, $$N=3500$$

Relative permeability of the core material, ${\mu }_r=800$

Magnetising current, $$I=1.2A$$

The magnetic field is given by the relation:

$B={\displaystyle \frac{{\mu }_0{\mu }_{r}lN}{2\pi r}}$

Where,

${\mu }_0=Permeabilityoffreespace=4\pi \times {10}^{-7}Tm{A}^{-1}$

$B={\displaystyle \frac{4\pi \times {10}^{-7}\times 800\times 1.2\times 3500}{2\pi \times 0.15}}=4.48T$

Thus, the magnetic field in the core is $4.48T$.

10. At a certain location in Africa, a compass points $12{}^{\circ }$ west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points above the horizontal. The horizontal component of the earth's field is measured to be 0.16 G. Specify the direction and magnitude of the earth's field at the location.

Ans: In the above question it is given that:

Angle of declination, $\theta =12{}^{\circ }$

Angle of dip, $\delta ={60}^0$

Horizontal component of earth's magnetic field, $$B_H=0.16G$$

Earth's magnetic field at the given location $$=B$$

We can relate B and $B_H$ as:

$B_H=B\cos \delta$

$\Rightarrow B={\displaystyle \frac{B_H}{\cos \delta }}$$={\displaystyle \frac{0.16}{\cos {60}^0}}=0.32G$

Earth's magnetic field lies in the vertical plane, $12{}^{\circ }$ West of the geographic meridian, making an angle of $$60{}^{\circ }$$ (upward) with the horizontal direction. Its magnitude is $0.32G$.

11. A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at $$22{}^{\circ }$$with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be$$0.35G$$. Determine the magnitude of the earth’s magnetic field at the place.

Ans: It is provided that,

The horizontal component of earth’s magnetic field, $$B_H=0.35G$$

The angle made by the needle with the horizontal plane (angle of dip)$$=\delta =22{}^{\circ }$$. 

Earth’s magnetic field strength is $$B$$. 

We can relate $$B$$ and $$B_H$$ as: $$B_H=B\cos \delta$$

$$\Rightarrow B={\displaystyle \frac{B_H}{\cos \delta }}$$

$$\Rightarrow B={\displaystyle \frac{0.35}{\cos 22{}^{\circ }}}=0.377G$$

Clearly, the strength of earth’s magnetic field at the given location is$$0.377G$$.

12. If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of $$0.25T$$ is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of $$30{}^{\circ }$$ with the direction of applied field? 

Ans: Given is the magnetic field strength, $$B=0.25T$$

Magnetic moment, $$M=0.6/T$$

The angle, $$\theta$$ between the axis of the turns of the solenoid and the direction of the external applied field is $$30{}^{\circ }$$ .

Hence, the torque acting on the solenoid is given as: 

$$\tau =MB\sin (\theta )$$

$$\Rightarrow \tau =0.6\times 0.25\sin (30{}^{\circ })$$

$$\Rightarrow \tau =7.5\times {10}^{-2}J$$

Hence the magnitude of torque is $$=7.5\times {10}^{-2}J$$.

13. A closely wound solenoid of $$800$$turns and area of cross section $$2.5\times {10}^{-4}{m}^2$$carries a current of $$3.0A$$. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment? 

Ans:  It is provided that the number of turns in the solenoid, $$n=800$$.

Area of cross-section, $$A=2.5\times {10}^{-4}{m}^2$$

Current in the solenoid, $$I=3.0A$$

A current-carrying solenoid is analogous to a bar magnet because a magnetic field develops along its axis, i.e., along its length joining the north and south poles.

The magnetic moment due to the given current-carrying solenoid is calculated as:

$$M=nIA=800\times 3\times 2.5\times {10}^{-4}=0.6J/T$$

Thus, the associated magnetic moment $$=0.6J/T$$

14. A short bar magnet placed with its axis at ${30}^0$ with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to $4.5\times {10}^{-2}J$. What is the magnitude of magnetic moment of the magnet?

Ans:  In the above question it is given that:

Magnetic field strength, $$B=0.25T$$

Torque on bar magnet, $T=4.5\times {10}^{-2}J$

Angle between the bar magnet external magnetic field, , $\theta ={30}^0$

We know that:

$$T=MB\sin \theta$$

$\Rightarrow M={\displaystyle \frac{T}{B\sin \theta }}$

$\Rightarrow M={\displaystyle \frac{4.5\times {10}^{-2}}{0.25\sin {30}^0}}=0.35J/T$

Hence, the magnetic moment is $0.35J/T$.

Short Answer Questions (3 Marks)

1. A short bar magnet of magnetic moment 0.9 J/T is placed with its axis at $$\mathbf{6}{\mathbf{0}}^{\mathbf{o}}$$ to a uniform magnetic field. It experiences a torque of 0.063 Nm. 

(i). calculate the strength of the magnetic field 

Ans: As $$\tau =MB\sin \theta$$,

$\theta ={60}^0$

$\tau =0.063Nm$

$M=0.9J/T$

$B={\displaystyle \frac{\tau }{M\sin \theta }}={\displaystyle \frac{0.063}{0.9\sin {60}^0}}=0.081T$

(ii). What orientation of the bar magnet corresponds to the equilibrium position in the magnetic field?

Ans: The magnet will be in stable equilibrium in the magnetic field if $\tau =0$

$\Rightarrow MB\sin 0^0=0$ i.e., When the magnet aligns itself parallel to the field.

2. A beam of electrons is moving with a velocity of $3\times {10}^{6}m/s$ and carries a current of $$1\mu A$$.

(a). How many electrons per second pass a given point?

Ans: We have $$I=1\mu A={10}^{-6}A$$

And $n={\displaystyle \frac{I}{q}}={\displaystyle \frac{{10}^{-6}}{1.6\times {10}^{-19}}}=6.25\times {10}^{12}$

(b). How many electrons are in 1m of the beam?

Ans: We know that electrons traverse a distance of $3\times {10}^{6}m$per second.

Thus, number of electrons in one meter of the beam$={\displaystyle \frac{6.25\times {10}^{12}}{3\times {10}^6}}=2.08\times {10}^6{m}^{-1}$

(c). What is the total force on all the electrons in 1m of the beam if it passes through the field of $0.1N{A}^{-1}{m}^{-1}$?

Ans: Force on one meter of the beam of electrons will be:

$F={\displaystyle \frac{0.1}{{10}^{-6}}}={10}^{5}N$

3. What is the main function of soft iron core used in a moving coil galvanometer? A galvanometer gives full deflection for Ig. Can it be converted into an ammeter of range $$I<Ig$$?

Ans: Soft iron core is used in the moving coil galvanometer because it increases the strength of the magnetic field thus increases the sensitivity of the galvanometer.

We know that, $$S={\displaystyle \frac{G{I}_g}{I-I_g}}$$

For $$I<Ig$$, S becomes negative. Clearly, it cannot be converted into an ammeter of range I < Ig.

4. Two wires loops PQRSP formed by joining two semi-circular wires of different radii carry a current I as shown in the figure. What is the direction of the magnetic induction at the centre C.?

Ans: Magnetic field due to semicircle QR at C is

$B_1={\displaystyle \frac{1}{2}}{\displaystyle \frac{{\mu }_0}{4\pi }}{\displaystyle \frac{2\pi I}{R_1}}$

Magnetic field due to semicircle is at C is

$B_2={\displaystyle \frac{1}{2}}{\displaystyle \frac{{\mu }_0}{4\pi }}{\displaystyle \frac{2\pi I}{R_2}}$

Net field,

$B=B_1-B_2$

$\Rightarrow B={\displaystyle \frac{2\pi I}{2}}{\displaystyle \frac{{\mu }_0}{4\pi }}({\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}})$

$\Rightarrow B={\displaystyle \frac{{\mu }_{0}I}{4}}({\displaystyle \frac{1}{R_1}}-{\displaystyle \frac{1}{R_2}})$

5. A circular coil is placed in uniform magnetic field of strength 0.10T normal to the plane of coil. If current in the coil is 5.0A. Find.

(a) Total torque on the coil

Ans: We have:

$$B=0.10T$$

$\theta =0^0$(Normal to plane of the coil)

$$I=5.0A$$,$$Area={10}^{-5}{m}^2$$, $n={10}^{29}/m^3$

$$\tau =MB\sin \theta =MB\sin 0^0=0$$, which is the required torque.

(b) Total force on the coil 

Ans: Total force on the coil $=0N$

(c). Average force on each electron due to magnetic field (The coil is made of copper wire of cross- sectional area and free electron density in copper is ${10}^{29}/m^3$)

Ans: We know that:

$F_{av}=q({\overrightarrow{v}}_d\times \overrightarrow{B})$

$$(I=neA{v}_d)$$

$\Rightarrow F_{av}=q({\displaystyle \frac{qI}{neA}}\times B)$

$\Rightarrow F_{av}={\displaystyle \frac{IB}{nA}}={\displaystyle \frac{5\times 0.10}{{10}^{29}\times {10}^{-5}}}=5\times {10}^{-25}N$

6. Using Ampere’s circuital law, derive an expression for magnetic field along the axis of a Toroidal solenoid.

Ans: If n is the number of turns per unit length; I be the current flowing through the Toroid;

Then from Ampere’s circuital law

$\oint \overrightarrow{B}.d\overrightarrow{l}={\mu }_0\times tot\text{al current flowing through tor}oid$

$\Rightarrow \oint \overrightarrow{B}.d\overrightarrow{l}={\mu }_0\left(2\pi rnI\right)$

$\Rightarrow \underset{0}{\overset{2\pi r}{\int }}Bdl\cos 0^0={\mu }_0\left(2\pi rnl\right)$

$\Rightarrow B\underset{0}{\overset{2\pi r}{\int }}dl={\mu }_0\left(2\pi rnI\right)$

$\Rightarrow B.2\pi r={\mu }_0\left(2\pi rnI\right)$

$\Rightarrow B={\mu }_{0}nI$, which is the required magnetic field here.

7. A short bar magnet of magnetic moment $$M=0.32J/T$$ is placed in a uniform magnetic field of $$0.15T$$. If the bar is free to rotate in the plane of the field, which orientation and would correspond to its

(a) Stable equilibrium? What is the potential energy of the magnet in this case? 

Ans: It is provided that moment of the bar magnet, $$M=0.32J/T$$.

External magnetic field, $$B=0.15T$$

It is considered as being in stable equilibrium, when the bar magnet is aligned along the magnetic field. Therefore, the angle $$\theta$$, between the bar magnet and the magnetic field is $$0{}^{\circ }$$ .

Potential energy of the system $$=-MB\cos (\theta )$$

$$\Rightarrow -MB\cos (\theta )=-0.32\times 0.15\times \cos (0)=-4.8\times {10}^{-2}J$$ 

Hence the potential energy is $$=-4.8\times {10}^{-2}J$$.

(b) Unstable equilibrium? What is the potential energy of the magnet in this case? 

Ans: It is provided that moment of the bar magnet, $$M=0.32J/T$$

External magnetic field, $$B=0.15T$$

When the bar magnet is aligned opposite to the magnetic field, it is considered as being in unstable equilibrium, $$\theta =180{}^{\circ }$$

Potential energy of the system is hence$$=-MB\cos (\theta )$$

$$\Rightarrow -MB\cos (\theta )=-0.32\times 0.15\times \cos (180{}^{\circ })=4.8\times {10}^{-2}J$$ 

Hence the potential energy is $$=4.8\times {10}^{-2}J$$.

8. A closely wound solenoid of $$2000$$ turns and area of cross-section $$1.6\times {10}^{-4}{m}^2$$, carrying a current of $$4.0A$$, is suspended through its centre allowing it to turn in a horizontal plane. 

(a) What is the magnetic moment associated with the solenoid? 

Ans: Given is the number of turns on the solenoid, $$n=2000$$

Area of cross-section of the solenoid, $$A=1.6\times {10}^{-4}{m}^2$$

Current in the solenoid, $$I=4A$$

The magnetic moment inside the solenoid at the axis is calculated as:

$$M=nAI=2000\times 1.6\times {10}^{-4}\times 4=1.28A{m}^2$$

(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of $$7.5\times {10}^{-2}T$$ is set up at an angle of $$30{}^{\circ }$$ with the axis of the solenoid?

Ans: Provided that,

Magnetic field, $$B=7.5\times {10}^{-2}T$$

Angle between the axis and the magnetic field of the solenoid, $$\theta =30{}^{\circ }$$

Torque, $$\tau =MB\sin (\theta )$$

$$\Rightarrow \tau =1.28\times 7.5\times {10}^{-2}\sin (30{}^{\circ })$$

$$\Rightarrow \tau =4.8\times {10}^{-2}Nm$$

Given the magnetic field is uniform, and the force on the solenoid is zero. The torque on the solenoid is $$4.8\times {10}^{-2}Nm$$.

9. A circular coil of $$16$$turns and radius $$10cm$$carrying a current of $$0.75$$A rests with its plane normal to an external field of magnitude $$5.0\times {10}^{-2}T$$. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of $$2.0/s$$. What is the moment of inertia of the coil about its axis of rotation? 

Ans: It is provided that,

The number of turns in the given circular coil solenoid, $$N=16$$

Radius of the coil, $$r=10cm=0.1m$$

Cross-section of the coil, $$A=\pi r^2=\pi \times {(0.1)}^2{m}^2$$

Current in the coil, $$I=0.75A$$

Magnetic field strength, $$B=5.0\times {10}^{-2}T$$

Frequency of oscillations of the coil, $$\upsilon =2.0/\mathrm{s}$$

Therefore, magnetic moment, $$M=NAI=NI\pi r^2$$

$$\Rightarrow M=16\times 7.5\times \pi \times {0.1}^2$$

$$\Rightarrow M=0.3777J/T$$

Frequency is given by the relation:

$$\nu ={\displaystyle \frac{1}{2\pi }}\sqrt{{\displaystyle \frac{MB}{I}}}$$

where,

$$I=$$Moment of inertia of the coil

$$\Rightarrow I={\displaystyle \frac{MB}{4{\pi }^2{\nu }^2}}$$ 

$$\Rightarrow I={\displaystyle \frac{0.377\times 5\times {10}^{-2}}{4{\pi }^2\times 2^2}}$$

$$\Rightarrow I=1.19\times {10}^{-4}kg{m}^2$$

Clearly, the moment of inertia of the coil about its axis of rotation $$1.19\times {10}^{-4}kg{m}^2$$.

10. A short bar magnet has a magnetic moment of $$0.48J/T$$. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of $$10cm$$from the centre of the magnet on

(a) The axis, 

Ans: Provided that the magnetic moment of the given bar magnet, $$M$$is $$0.48J/T$$

Given distance, $$d=10cm=0.1m$$

The magnetic field at d-distance, from the centre of the magnet on the axis is given by the relation:

$$B={\displaystyle \frac{{\mu }_0}{4\pi }}{\displaystyle \frac{2M}{d^3}}$$

here,

$${\mu }_0=$$ Permeability of free space$$=4\pi \times {10}^{-7}Tm/A$$

Substituting these values, $$B$$ becomes as follows:

$$\Rightarrow B={\displaystyle \frac{4\pi \times {10}^{-7}}{4\pi }}{\displaystyle \frac{2\times 0.48}{{0.1}^3}}$$

$$\Rightarrow B=0.96\times {10}^{-4}T=0.96G$$

The magnetic field is $$0.96G$$ along the South-North direction.

(b). The equatorial lines (normal bisector) of the magnet.

Ans: The magnetic field at a point which is $$d=10cm=0.1m$$ away on the equatorial of the magnet is given as:

$$B={\displaystyle \frac{{\mu }_0}{4\pi }}{\displaystyle \frac{M}{d^3}}$$

$$\Rightarrow B={\displaystyle \frac{4\pi \times {10}^{-7}}{4\pi }}{\displaystyle \frac{0.48}{{0.1}^3}}$$

$$\Rightarrow B=0.48\times {10}^{-4}T=0.48G$$

The magnetic field is $$0.48G$$along the North-South direction.

11. A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at $$14cm$$ from the centre of the magnet. The earth’s magnetic field at the place is $$0.36G$$and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null–point (i.e., $$14cm$$) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field).

Ans: Provided that,

The magnetic field of Earth at the given place,$$H=0.36G$$

The magnetic field at a $$d$$-distance, on the axis of the magnet is given as:

$$B_1={\displaystyle \frac{{\mu }_0}{4\pi }}{\displaystyle \frac{2M}{d^3}}=H$$

Here,

$${\mu }_0=$$ Permeability of free space$$=4\pi \times {10}^{-7}Tm/A$$

$$M=$$ The magnetic moment

The magnetic field at the same distance $$d$$, on the equatorial line of the magnet is given as:

$$B_2={\displaystyle \frac{{\mu }_0}{4\pi }}{\displaystyle \frac{M}{d^3}}$$

$$\Rightarrow B_2=H/2$$ (comparing with $$B_1$$)

Therefore, the total magnetic field,

$$B=B_1+B_2$$

$$\Rightarrow B=H+H/2$$

$$\Rightarrow B=0.36+0.18=0.54$$

Clearly, the magnetic field is 0.54 G along the direction of earth’s magnetic field.

12. A long straight horizontal cable carries a current of $$2.5A$$ in the direction $$10{}^{\circ }$$ south of west to $$10{}^{\circ }$$ north of east. The magnetic meridian of the place happens to be $$10{}^{\circ }$$ west of the geographic meridian. The earth’s magnetic field at the location is $$0.33G$$, and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable). (At neutral points, magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth’s magnetic field.)

Ans: Provided that,

Current in the wire, $$I=2.5A$$

The angle of dip at the location, $$\delta =0{}^{\circ }$$

The Earth’s magnetic field, $$H=0.33G=0.33\times {10}^{-4}T$$

The horizontal component of earth’s magnetic field is given as:

$$H_H=H\cos \delta =0.33\times {10}^{-4}\times \cos 0{}^{\circ }=0.33\times {10}^{-4}T$$

The magnetic field at the neutral point at a distance R from the cable is given by the relation:

$$H_H={\displaystyle \frac{{\mu }_0}{2\pi }}{\displaystyle \frac{I}{R}}$$

here,  $${\mu }_0$$ = Permeability of free space $$=4\pi \times {10}^{-7}Tm/A$$

$$\Rightarrow R={\displaystyle \frac{{\mu }_0}{2\pi }}{\displaystyle \frac{I}{H_H}}$$

$$\Rightarrow R={\displaystyle \frac{4\pi \times {10}^{-7}}{2\pi }}{\displaystyle \frac{2.5}{0.33\times {10}^{-4}}}=15.15\times {10}^{-3}m=1.51cm$$

Clearly, a set of neutral points lie on a straight line parallel to the cable at a perpendicular distance of $$1.51cm$$ above the plane of the paper.

13. A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of $$45{}^{\circ }$$ with the magnetic meridian. When the current in the coil is$$0.35A$$, the needle points west to east.

(a) Determine the horizontal component of the earth’s magnetic field at the location. 

Ans: Provided that,

The number of turns in the given circular coil, $$N=30$$

The radius of the given circular coil, $$r=12cm=0.12m$$

Current in the coil, $$I=0.35A$$

Angle of dip, $$\delta =45{}^{\circ }$$

The magnetic field due to the current I, at a distance r, is given as:

$$B_{}=4{\displaystyle \frac{{\mu }_0}{2\pi }}{\displaystyle \frac{I}{r}}$$

here,

$${\mu }_0$$ = Permeability of free space $$=4\pi \times {10}^{-7}Tm/A$$

$$\Rightarrow B={\displaystyle \frac{4\pi \times {10}^{-7}}{4\pi }}{\displaystyle \frac{2\pi \times 30\times 0.35}{0.12}}$$

$$\Rightarrow B=54.9\times {10}^{-4}T=0.549G$$

The compass needle points West to East. Hence, the horizontal component of earth’s magnetic field is given as:

$$B_H=B.\sin \delta$$

$$\Rightarrow B_H=0.549\sin 45{}^{\circ }=0.388G$$

(b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of $$90{}^{\circ }$$ in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.

Ans: If the direction of the current flowing in the coil is reversed and if the coil is also rotated about its vertical axis by an angle of $$90{}^{\circ }$$, the needle will rearrange and reverse its original direction. In the given case, the needle would point from East to West.

14. A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is $$60{}^{\circ }$$, and one of the fields has a magnitude of $$1.2\times {10}^{-2}T$$. If the dipole comes to stable equilibrium at an angle of $$15{}^{\circ }$$with this field, what is the magnitude of the other field?

Ans: Provided that,

Magnitude of one of the magnetic fields, $$B_1=1.2\times {10}^{-2}T$$

Magnitude of the other magnetic field is $$B_2$$.

Angle between the above-mentioned two fields, $$\theta =60{}^{\circ }$$

At the state of stable equilibrium, the angle between the dipole and field$$B_1$$ is $${\theta }_1=15{}^{\circ }$$

Angle between the dipole and field $$B_2$$ is $${\theta }_2=\theta -{\theta }_1=60{}^{\circ }-15{}^{\circ }=45{}^{\circ }$$ 

At a rotational equilibrium, the torques experienced by the dipole, due to both the fields, must balance each other.

Therefore, torque due to field $$B_1=$$Torque due to field $$B_2$$

$$M{B}_1\sin {\theta }_1=M{B}_2\sin {\theta }_2$$

Where,

$$M=$$Magnetic moment of the dipole

$$\Rightarrow B_2={\displaystyle \frac{B_1\sin {\theta }_1}{\sin {\theta }_2}}$$

$$\Rightarrow B_2=4.39\times {10}^{-3}T$$

Clearly, the magnitude of the other magnetic field is $$4.39\times {10}^{-3}T$$.

15. The magnetic moment vectors $${\mu }_s$$and $${\mu }_l$$associated with the intrinsic spin angular momentum $$\overrightarrow{S}$$ and orbital angular momentum $$\overrightarrow{l}$$, respectively, of an electron are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by

$${\mu }_s=-(e/m)S$$

$${\mu }_l=-(e/2m)l$$

Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.

Ans: According to the definition of magnetic moment-$${\mu }_l$$ and orbital angular momentum-$$l$$.

Magnetic moment associated with the motion of the electron is:

$${\mu }_l=iA=-(e/T).\pi r^2$$

And the corresponding angular momentum is:

$$l=mvr=m(2\pi r/T)r^{}$$

Where $$r$$ is the radius of the orbit, which the mass of electron is $$m$$and its charge $$(-e)$$ completes in time $$T$$. 

Dividing $${\mu }_{l}byl$$, one would get:

$${\displaystyle \frac{{\mu }_l}{l}}={\displaystyle \frac{-e}{T}}.\pi r^2\times {\displaystyle \frac{T}{m^2\pi r^2}}=-{\displaystyle \frac{e}{2m}}$$

$$\Rightarrow {\overrightarrow{\mu }}_l=-\left({\displaystyle \frac{e}{2m}}\right)\overrightarrow{l}$$

Evidently, it can be seen that $${\overrightarrow{\mu }}_l$$ and $$\overrightarrow{l}$$ will be antiparallel (both being normal to the plane of the orbit). 

In contrast, $${\displaystyle \frac{{\mu }_s}{s}}=\left({\displaystyle \frac{e}{m}}\right)$$ and it is derived on the basis of quantum mechanics and is verified experimentally.

Long Answer Questions (5 Marks)

1. A particle of mass m and charge q moving with a uniform speed normal to a uniform magnetic field B describes a circular path of radius & Derive expressions for

(1) Radius of the circular path (2) time period of revolution (3) Kinetic energy of the particle?

Ans: A particle of mass (m) and change (q) moving with velocity normal to describes a circular path if

$${\displaystyle \frac{m{v}^2}{r}}=qBv\sin \theta$$

$$\Rightarrow {\displaystyle \frac{m{v}^2}{r}}=qBv(\because \theta =90{}^{\circ })$$

$\Rightarrow r={\displaystyle \frac{mv}{Bq}}$ …… (1)

This is the required radius of the circular path.

Now, since 

Time period of Revolution $$duringcircularpath={\displaystyle \frac{Circumf\text{erence of c}ircle}{velocity}}$$;

$T={\displaystyle \frac{2\pi r}{v}}$

$\Rightarrow T={\displaystyle \frac{2\pi r.m}{Bqr}}$ ……. (from 1)

$\Rightarrow T={\displaystyle \frac{2\pi m}{Bq}}$ …… (2)

This is the required time period.

Now,

$$\text{Kinetic energy K}\text{.E}={\displaystyle \frac{1}{2}}m{v}^2$$

$$\Rightarrow \text{K}\text{.E}={\displaystyle \frac{1}{2}}m{\left({\displaystyle \frac{Bqr}{m}}\right)}^2$$

$$\Rightarrow \text{K}\text{.E}={\displaystyle \frac{B^2{q}^2{r}^2}{2m}}$$, which is the required kinetic energy.

2. Write an expression for the force experienced by the charged particle moving in a uniform magnetic field B with the help of labeled diagram explain the working of cyclotron. Show that cyclotron frequency does not depend upon the speed of the particle.

Ans: Force experienced by the charged particle moving at right angles to uniform magnetic field $\overrightarrow{B}$with velocity $\overrightarrow{v}$ is given by $$F=q(\text{vec{v}}\times \text{vec{B}})$$ Initially Dee $$D_1$$ is negatively charged and Dee $$D_2$$ is positively charged so, the positive ion will get accelerated towards Dee $$D_1$$. Since the magnetic field is uniform and acting at right angles to the plane of the Dees so the ion completes a circular path in $$D_1$$when ions come out into the gap, polarity of the Dee’s gets reversed used the ion is further accelerated towards Dee $$D_2$$ with greater speed and cover a bigger semi-circular path. This process is separated time and again and the speed of the ion becomes faster till it reaches the periphery of the dees where it is brought out by means of a deflecting plate and is made to bombard the target.

Since $$F=qVBsin{90}^0$$ provides the necessary centripetal force to the ion to cover a circular path so we can say ${\displaystyle \frac{m{v}^2}{r}}=qvB$

$r={\displaystyle \frac{mv}{Bq}}$ …… (1)

Time period = ${\displaystyle \frac{2\pi r}{v}}={\displaystyle \frac{2\pi rm}{Bqr}}={\displaystyle \frac{2\pi m}{Bq}}$

$V={\displaystyle \frac{1}{T}}={\displaystyle \frac{Bq}{2\pi m}}$

Thus, frequency is independent of velocity.

3. 

(a) Obtain an expression for the torque acting on a current carrying circular loop.

Ans: ABCD is a square loop of length (L) and area (A). Let I be the current flowing in the anticlockwise direction. Let $\theta$ be the angle between the normal to the loop and magnetic field $\overrightarrow{B}$.

Force acting on arm AB of the loop

${\overrightarrow{F}}_1=I(\overrightarrow{L}\times \overrightarrow{B})\left(outwards\right)$

Force on arm CD

${\overrightarrow{F}}_2=I(\overrightarrow{L}\times \overrightarrow{B})\left(inwards\right)$

Force on arm BC

${\overrightarrow{F}}_3=I(\overrightarrow{L}\times \overrightarrow{B})\left(downwards\right)$

Force on arm DA

${\overrightarrow{F}}_4=I(\overrightarrow{L}\times \overrightarrow{B})\left(upwards\right)$

Since ${\overrightarrow{F}}_3$ and ${\overrightarrow{F}}_4$ are equal and opposite and also acts along the same line, hence they cancel each other.

${\overrightarrow{F}}_1$ and ${\overrightarrow{F}}_2$ are also equal and opposite but their line of action is different, so they form a couple and makes the rectangular loop rotate anti clockwise.

Thus $$\tau =eitherforce\times \mathrm{\perp }distance$$

$\tau =I(\overrightarrow{L}\times \overrightarrow{B})\times DN$

$\tau =I(\overrightarrow{L}\times \overrightarrow{B})\times b\sin \theta$

$\tau =ILB\sin {90}^0\times b\sin \theta$

$\tau =IAB\sin \theta$

For loop of N turns

$\tau =NIAB\sin \theta$

$\tau =MB\sin \theta (\because M=NIA)$

$\overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}$

Where M is magnetic moment of the loop.

(b). What is the maximum torque on a galvanometer coil 5 cm $\times$ 12 cm of 600 turns when carrying a current of  $$\mathbf{1}{\mathbf{0}}^{-\mathbf{5}}$$ A in a field where flux density is $0.10Wb/m^2$?

Ans: It is known that

$\tau =NIAB\sin \theta$

Torque will be maximum when $$\theta ={90}^o$$

${\tau }_{max}=NIAB(\because \sin {90}^0=1)$

${\tau }_{max}=600\times {10}^{-5}\times \left(0.10\right)(60\times {10}^{-4})=3.6\times {10}^{-6}Nm$.

4. The current sensitivity of a moving coil galvanometer increases by 20% when its resistance is increased by a factor of two. Calculate by what factor, the voltage sensitivity changes?

Ans: We know that:

Current sensitivity, ${\displaystyle \frac{\alpha }{I}}={\displaystyle \frac{nBA}{k}}$ …… (i) 

Voltage sensitivity, ${\displaystyle \frac{\alpha }{V}}={\displaystyle \frac{nBA}{kr}}$ …… (ii)

Resistance of a galvanometer increases when n and A are changed

Given $$R^{\prime }=2R$$

Then $$n=n^{\prime }$$ and $$A=A^{\prime }$$

New current sensitivity

${\displaystyle \frac{{\alpha }^{\prime }}{I^{\prime }}}={\displaystyle \frac{n^{\prime }B{A}^{\prime }}{k}}$ …… (iii)

New voltage sensitivity

${\displaystyle \frac{{\alpha }^{\prime }}{V}}={\displaystyle \frac{{\alpha }^{\prime }}{I^{\prime }{R}^{\prime }}}={\displaystyle \frac{n^{\prime }B{A}^{\prime }}{2kr}}$ …… (iv)

Since, ${\displaystyle \frac{{\alpha }^{\prime }}{I^{\prime }}}={\displaystyle \frac{120\alpha }{100I}}$ …… (v)

From (i) and (iii)

${\displaystyle \frac{n^{\prime }{A}^{\prime }B}{R}}={\displaystyle \frac{\alpha 120}{I100}}$

${\displaystyle \frac{n^{\prime }{A}^{\prime }B}{R}}={\displaystyle \frac{nAB120}{k100}}$

$n^{\prime }{A}^{\prime }={\displaystyle \frac{6}{5}}nA$

Using equation (iv)

${\displaystyle \frac{{\alpha }^{\prime }}{V}}={\displaystyle \frac{6}{5}}{\displaystyle \frac{nBA}{2kr}}={\displaystyle \frac{3}{5}}{\displaystyle \frac{\alpha }{V}}$

Thus, voltage sensitivity decreases by a factor of ${\displaystyle \frac{3}{5}}$.

5. 

(a) Show how a moving coil galvanometer can be converted into an ammeter.

Ans: A galvanometer can be converted into an ammeter by connecting a low resistance called shunt parallel to the galvanometer. Since G and $$R_S$$ are in parallel voltage across then is same $I_g{R}_G=(I-I_g)R_s$.

$R_s=\left({\displaystyle \frac{I_g}{I-I_g}}\right)R_G$

(b) A galvanometer has a resistance 30 and gives a full-scale deflection for a current of 2mA. How much resistance in what way must be connected to convert into?

(i). An ammeter of range 0.3A

Ans: We have:

$$I=0.3A$$, $$G=30\mathrm{\Omega }$$, $$Ig=2mA=2\times {10}^{-3}A$$

$Shunt\left(S\right)={\displaystyle \frac{I_{g}G}{I-I_g}}$

$S={\displaystyle \frac{2\times {10}^{-3}\times 30}{(0.3-2\times {10}^{-3})}}=0.2\mathrm{\Omega }$

(ii). A voltmeter of range 0.2V.

Ans: We have:

$$G=30\mathrm{\Omega }$$, $$Ig=2mA=2\times {10}^{-3}A$$, $$V=0.2V$$

$Shu\text{nt Re}sis\tan ce\left(R\right)=({\displaystyle \frac{V}{I_g}}-G)$

$\Rightarrow \left(R\right)=({\displaystyle \frac{0.2}{2\times {10}^{-3}}}-30)=70\mathrm{\Omega }$

6. A monoenergetic ($$18keV$$) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of $$0.04G$$normal to the initial direction. Estimate the up or down deflection of the beam over a distance of $$30cm(m_e=9.11\times {10}^{-31}kg)$$. (Note: Data in this exercise are so chosen that the answer will give you an idea of the effect of earth’s magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.)

Ans: Provided that,

Energy of an electron beam, $$E=18keV=18\times {10}^{3}eV$$

Charge on an electron, $$e=1.6\times {10}^{-19}C$$

$$\therefore E=18\times {10}^3\times 1.6\times {10}^{-19}V=2.88\times {10}^{-15}V$$

The magnetic field, $$B=0.04G$$

The mass of an electron, $$m_e=9.11\times {10}^{-31}kg$$

Distance till where the electron beam travels, $$d=30cm=0.3m$$

We can write the kinetic energy carried by the electron beam as:

$$E={\displaystyle \frac{1}{2}}m{v}^2$$

$$\Rightarrow v=\sqrt{{\displaystyle \frac{2E}{m}}}$$

$$\Rightarrow v=\sqrt{2\times {\displaystyle \frac{2.88\times {10}^{-15}}{9.11\times {10}^{-31}}}}=0.795\times {10}^{8}m/s$$

The electron beam deflects and gets in a circular path of radius, $$r$$ .

The force experienced due to the magnetic field balances the centripetal force of the path.

$$BeV={\displaystyle \frac{m{v}^2}{r}}$$

$$\Rightarrow r={\displaystyle \frac{mv}{Be}}$$

$$\Rightarrow r={\displaystyle \frac{9.11\times {10}^{-31}\times 0.795\times {10}^8}{0.4\times {10}^{-4}\times 1.6\times {10}^{-19}}}=11.3m$$

Let the up-down deflection of the beam be $$x=r(1-\cos \theta )$$.

Where, 

$$\theta$$ is the angle of declination given by,

$$\theta ={\sin }^{-1}(d/r)=1.521{}^{\circ }$$