Class 12th Physics Chapter-4 Solutions

Question 4.1

A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field $B$ at the centre of the coil?

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Given

• Number of turns, $N=100$

• Radius of coil, $R=8.0\text{ cm}=0.08\text{ m}$
  

• Current, $I=0.40\text{ A}$

• Permeability of free space,

  $${\mu }_0=4\pi \times {10}^{-7}{\text{T m A}}^{-1}$$

Formula Used

For a circular coil with $N$ turns, the magnetic field at the centre is:

$$B=\frac{{\mu }_{0}NI}{2R}$$

(This result follows from the Biot–Savart law for a circular current loop, as discussed in the NCERT text)

Substitution

$$B=\frac{(4\pi \times {10}^{-7})\times 100\times 0.40}{2\times 0.08}$$

$$B=\frac{160\pi \times {10}^{-7}}{0.16}$$

$$B=1000\pi \times {10}^{-7}\text{ T}$$

$$B=\pi \times {10}^{-4}\text{ T}$$

Final Answer

$$\boxed{{\displaystyle B\approx 3.14\times {10}^{-4}\text{ T}}}$$

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Question 4.2

A long straight wire carries a current of 35 A. What is the magnitude of the magnetic field $B$ at a point 20 cm from the wire?

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Given

• Current in the wire, $I=35\text{ A}$

• Distance from the wire,

  $$r=20\text{ cm}=0.20\text{ m}$$

• Permeability of free space,

  $${\mu }_0=4\pi \times {10}^{-7}{\text{T m A}}^{-1}$$

Formula Used

For a long straight current-carrying wire, the magnetic field at a distance $r$ is:

$$B=\frac{{\mu }_{0}I}{2\pi \mathrm{r<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"></span>}}$$

(This expression follows from Ampere’s circuital law / Biot–Savart law as given in NCERT Physics, Moving Charges and Magnetism)

Substitution

$$B=\frac{4\pi \times {10}^{-7}\times 35}{2\pi \times 0.20}$$

Cancel $\pi$:

$$B=\frac{4\times 35\times {10}^{-7}}{0.40}$$

$$B=\frac{140\times {10}^{-7}}{0.40}$$

$$B=350\times {10}^{-7}\text{ T}$$

$$B=3.5\times {10}^{-5}\text{ T}$$

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Question 4.3

A long straight wire in the horizontal plane carries a current of 50 A in the north-to-south direction. Give the magnitude and direction of the magnetic field $B$ at a point 2.5 m east of the wire.

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Given

• Current, $I=50\text{ A}$

• Distance from the wire, $r=2.5\text{ m}$

• Direction of current: North → South

• Permeability of free space:

  $${\mu }_0=4\pi \times {10}^{-7}{\text{T m A}}^{-1}$$

Magnitude of Magnetic Field

For a long straight current-carrying wire,

$$B=\frac{{\mu }_{0}I}{2\pi r}$$

(Substantiated by Ampere’s circuital law / Biot–Savart law as discussed in NCERT Physics, Moving Charges and Magnetism)

Substitution

$$B=\frac{4\pi \times {10}^{-7}\times 50}{2\pi \times 2.5}$$Cancel $\pi$:

$$B=\frac{200\times {10}^{-7}}{5}$$

$$B=40\times {10}^{-7}\text{ T}$$

$$\boxed{{\displaystyle B=4.0\times {10}^{-6}\text{ T}}}$$

Direction of Magnetic Field

Use the right-hand thumb rule:

• Point the right-hand thumb in the direction of current (north to south).

• The curl of the fingers gives the direction of magnetic field lines.

At a point east of the wire, the curled fingers point vertically upward, i.e., out of the horizontal plane.

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Question 4.4

A horizontal overhead power line carries a current of 90 A in the east-to-west direction. What is the magnitude and direction of the magnetic field due to the current at a point 1.5 m below the line?

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Given

• Current, $I=90\text{ A}$

• Distance from the wire, $r=1.5\text{ m}$

• Direction of current: East → West

• Permeability of free space:

  $${\mu }_0=4\pi \times {10}^{-7}{\text{T m A}}^{-1}$$

Magnitude of Magnetic Field

For a long straight current-carrying wire,

$$B=\frac{{\mu }_{0}I}{2\pi r}$$

(as derived using Ampere’s circuital law / Biot–Savart law in NCERT Physics, Moving Charges and Magnetism )

Substitution

$$B=\frac{4\pi \times {10}^{-7}\times 90}{2\pi \times 1.5}$$

Cancel $\pi$:

$$B=\frac{360\times {10}^{-7}}{3}$$

$$B=120\times {10}^{-7}\text{ T}$$

$$\boxed{{\displaystyle B=1.2\times {10}^{-5}\text{ T}}}$$

Direction of Magnetic Field

Apply the right-hand thumb rule:

• Thumb → direction of current (east to west)

• Curled fingers → direction of magnetic field lines around the wire

At a point below the wire, the magnetic field is directed towards the south.

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Question 4.5

What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of ${30}^{\circ }$ with the direction of a uniform magnetic field of 0.15 T?

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Given

• Current in the wire, $I=8\text{ A}$

• Magnetic field, $B=0.15\text{ T}$

• Angle between current and magnetic field,

  $$\theta ={30}^{\circ }$$

Formula Used

The magnetic force on a current-carrying conductor of length $l$ in a uniform magnetic field is:

$$F=IlB\sin \theta$$

Therefore, force per unit length is:

$$\frac{F}{l}=IB\sin \theta$$

(This relation follows from the Lorentz force law as discussed in NCERT Physics, Moving Charges and Magnetism)

Substitution

$$\frac{F}{l}=8\times 0.15\times \sin {30}^{\circ }$$

$$\sin {30}^{\circ }=\frac{1}{2}$$

$$\frac{F}{l}=8\times 0.15\times 0.5$$

$$\frac{F}{l}=0.6{\text{N m}}^{-1}$$

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Question 4.6

A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is 0.27 T. What is the magnetic force on the wire?

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Given

• Current in the wire, $I=10\text{ A}$

• Length of wire,

  $$l=3.0\text{ cm}=0.03\text{ m}$$

• Magnetic field inside solenoid,

  $$B=0.27\text{ T}$$

• Angle between current and magnetic field,

  $$\theta ={90}^{\circ }(\text{wire is perpendicular to solenoid axis})$$

Formula Used

Magnetic force on a current-carrying conductor in a uniform magnetic field:

$$F=IlB\sin \theta$$

(This follows from the Lorentz force law for a current-carrying conductor, as discussed in NCERT Physics, Moving Charges and Magnetism )

Substitution

$$F=10\times 0.03\times 0.27\times \sin {90}^{\circ }$$

$$\sin {90}^{\circ }=1$$

$$F=10\times 0.03\times 0.27$$

$$F=0.081\text{ N}$$

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Question 4.7

Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

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Given

• Current in wire A, $I_A=8.0\text{ A}$

• Current in wire B, $I_B=5.0\text{ A}$

• Separation between wires,

  $$d=4.0\text{ cm}=0.04\text{ m}$$

• Length of wire A considered,

  $$l=10\text{ cm}=0.10\text{ m}$$

• Permeability of free space,

  $${\mu }_0=4\pi \times {10}^{-7}{\text{T m A}}^{-1}$$

Formula Used

Force between two long parallel current-carrying conductors:

$$F=\frac{{\mu }_0{I}_A{I}_{B}l}{2\pi d}$$

(This relation is derived using the magnetic field due to a straight wire and the Lorentz force law, as discussed in NCERT Physics, Moving Charges and Magnetism )

Substitution

$$F=\frac{4\pi \times {10}^{-7}\times 8.0\times 5.0\times 0.10}{2\pi \times 0.04}$$

Cancel $\pi$:

$$F=\frac{4\times {10}^{-7}\times 8\times 5\times 0.10}{2\times 0.04}$$

$$F=\frac{16\times {10}^{-7}}{0.08}$$

$$F=200\times {10}^{-7}\text{ N}$$

$$F=2.0\times {10}^{-5}\text{ N}$$

Direction of Force

Since the currents in the two wires are in the same direction, the wires attract each other.

Force on wire A is directed towards wire B.

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Question 4.8

A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of the magnetic field $B$ inside the solenoid near its centre.

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Given

• Length of solenoid,

  $$l=80\text{ cm}=0.80\text{ m}$$

• Number of layers = 5

• Turns per layer = 400

$$\Rightarrow \text{Total turns }N=5\times 400=2000$$

• Current,

  $$I=8.0\text{ A}$$

• Diameter = 1.8 cm ⇒ Radius = 0.9 cm = 0.009 m

• Permeability of free space,

  $${\mu }_0=4\pi \times {10}^{-7}{\text{T m A}}^{-1}$$

Since the solenoid is long compared to its diameter, the field near the centre is uniform.

Formula Used

Magnetic field inside a long solenoid:

$$B={\mu }_{0}nI$$

where

$$n=\frac{N}{l}\text{(number of turns per unit length)}$$

(This result follows from Ampere’s circuital law as discussed in NCERT Physics, Moving Charges and Magnetism)

Calculation

Turns per unit length:

$$n=\frac{2000}{0.80}=2500{\text{m}}^{-1}$$

Magnetic field:

$$B=(4\pi \times {10}^{-7})\times 2500\times 8$$

$$B=4\pi \times {10}^{-7}\times 20000$$

$$B=8\pi \times {10}^{-3}\text{ T}$$

$$B\approx 2.5\times {10}^{-2}\text{ T}$$

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Question 4.9

A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of ${30}^{\circ }$ with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

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Given

• Side of square coil,

  $$a=10\text{ cm}=0.10\text{ m}$$

• Number of turns,

  $$N=20$$

• Current,

  $$I=12\text{ A}$$

• Magnetic field,

  $$B=0.80\text{ T}$$

• Angle between normal to the coil and magnetic field,

  $$\theta ={30}^{\circ }$$

Formula Used

Torque on a current-carrying coil in a uniform magnetic field:

$$\tau =NIAB\sin \theta$$

where

$$A=\text{area of the coil}$$

(This expression follows from the magnetic dipole moment of a current loop, as discussed in NCERT Physics, Moving Charges and Magnetism )

Calculation

Area of square coil:

$$A=a^2=(0.10{)}^2=0.01{\text{m}}^2$$

Substitute values:

$$\tau =20\times 12\times 0.01\times 0.80\times \sin {30}^{\circ }$$

$$\sin {30}^{\circ }=\frac{1}{2}$$

$$\tau =20\times 12\times 0.01\times 0.80\times 0.5$$

$$\tau =0.96\text{ N m}$$

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Question 4.10

Two moving coil meters, $M_1$ and $M_2$, have the following particulars:

• Meter $M_1$:
  $R_1=10\mathrm{\Omega },N_1=30,A_1=3.6\times {10}^{-3}{\text{m}}^2,B_1=0.25\text{ T}$

• Meter $M_2$:
  $R_2=14\mathrm{\Omega },N_2=42,A_2=1.8\times {10}^{-3}{\text{m}}^2,B_2=0.50\text{ T}$

Compare the current sensitivities of the two meters. Which one is more sensitive?

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Concept Used: Current Sensitivity of a Moving Coil Meter

For a moving coil galvanometer, current sensitivity is defined as angular deflection per unit current:

$$\text{Current sensitivity}\propto \frac{NAB}{k}$$

where

• $N$ = number of turns

• $A$ = area of the coil

• $B$ = magnetic field

• $k$ = torsional constant of the spring

 If the meters are of similar construction, we may assume $k$ is the same for both.
Hence, sensitivity depends on $NAB$.

(This result follows from the torque on a current loop in a magnetic field, discussed in NCERT Physics, Moving Charges and Magnetism )

Calculation

For meter $M_1$:

$$(NAB{)}_1=30\times (3.6\times {10}^{-3})\times 0.25$$

$$(NAB{)}_1=30\times 0.9\times {10}^{-3}$$

$$(NAB{)}_1=27\times {10}^{-3}$$

For meter $M_2$:

$$(NAB{)}_2=42\times (1.8\times {10}^{-3})\times 0.50$$

$$(NAB{)}_2=42\times 0.9\times {10}^{-3}$$

$$(NAB{)}_2=37.8\times {10}^{-3}$$

Comparison

$$\frac{\text{Sensitivity of }{M}_2}{\text{Sensitivity of }{M}_1}=\frac{37.8}{27}\approx 1.4$$

Ratio of voltage sensitivity

$$\text{Voltage sensitivity}=\frac{\text{current sensitivity}}{R}$$

$$\Rightarrow \text{Voltage sensitivity}\propto \frac{NAB}{R}$$
$$\frac{S_{V2}}{S_{V1}}=\frac{(NAB{)}_2\mathrm{/}{R}_2}{(NAB{)}_1\mathrm{/}{R}_1}=\frac{37.8\mathrm{/}14}{27\mathrm{/}10}$$

$$\frac{S_{V2}}{S_{V1}}=\frac{37.8\times 10}{27\times 14}=1$$

Final Answer

• Meter $M_2$ is more sensitive than meter $M_1$.

• $M_2$ is approximately 1.4 times more sensitive than $M_1$.

• Voltage Sensitivity = 1

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Question 4.11

In a chamber, a uniform magnetic field of 6.5 G is maintained. An electron is shot into the field with a speed of $4.8\times {10}^6{\text{m s}}^{-1}$ normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit.

Given:

$$B=6.5\text{ G}=6.5\times {10}^{-4}\text{ T},v=4.8\times {10}^6{\text{m s}}^{-1}$$

$$e=1.5\times {10}^{-19}\text{ C},m_e=9.1\times {10}^{-31}\text{ kg}$$

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Why is the path circular?

When a charged particle moves perpendicular to a uniform magnetic field, it experiences a magnetic (Lorentz) force:

$$\vec{F}=q\vec{v}\times \vec{B}$$

• The force is always perpendicular to the velocity of the electron.

• Hence, the force does no work (speed remains constant).

• A force perpendicular to velocity acts as a centripetal force, continuously changing only the direction of motion.

Therefore, the electron moves in a circular path.

Radius of the circular orbit

For circular motion:

$$\text{Centripetal force}=\text{Magnetic force}$$
$$\frac{m{v}^2}{r}=evB$$

$$r=\frac{mv}{eB}$$

Substitution

$$r=\frac{(9.1\times {10}^{-31})(4.8\times {10}^6)}{(1.5\times {10}^{-19})(6.5\times {10}^{-4})}$$

Numerator:

$$9.1\times 4.8\times {10}^{-25}=43.68\times {10}^{-25}$$

Denominator:

$$1.5\times 6.5\times {10}^{-23}=9.75\times {10}^{-23}$$

$$r=\frac{43.68}{9.75}\times {10}^{-2}$$

$$r\approx 4.48\times {10}^{-2}\text{ m}$$

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Question 4.12

In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

Given (from Q. 4.11):

$$B=6.5\text{ G}=6.5\times {10}^{-4}\text{ T},e=1.5\times {10}^{-19}\text{ C},m_e=9.1\times {10}^{-31}\text{ kg}$$

Frequency of revolution

For a charged particle moving perpendicular to a uniform magnetic field, the cyclotron (revolution) frequency is:

$$f=\frac{eB}{2\pi m}$$

Substitution

$$f=\frac{(1.5\times {10}^{-19})(6.5\times {10}^{-4})}{2\pi (9.1\times {10}^{-31})}$$

First compute:

$$\frac{eB}{m}=\frac{9.75\times {10}^{-23}}{9.1\times {10}^{-31}}\approx 1.07\times {10}^8{\text{s}}^{-1}$$

$$f=\frac{1.07\times {10}^8}{2\pi }\approx 1.7\times {10}^7\text{ Hz}$$

$$\boxed{{\displaystyle f\approx 1.7\times {10}^7\text{ Hz}}}$$

Does the frequency depend on the speed of the electron?

No.
From the expression

$$f=\frac{eB}{2\pi m}$$

the frequency depends only on:

• charge of the particle ($e$),

• magnetic field ($B$),

• mass of the particle ($m$).

It is independent of the speed (or energy) of the electron, as long as the speed is non-relativistic.

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Question 4.13

(a)

A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of ${60}^{\circ }$ with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.

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Given

• Number of turns, $N=30$

• Radius, $r=8.0\text{ cm}=0.08\text{ m}$

• Current, $I=6.0\text{ A}$

• Magnetic field, $B=1.0\text{ T}$

• Angle between field and normal to the coil, $\theta ={60}^{\circ }$

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Formula Used

Torque on a current-carrying coil in a uniform magnetic field:

$$\tau =NIAB\sin \theta$$

where $A$ is the area of the coil.

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Calculation

Area of the circular coil:

$$A=\pi r^2=\pi (0.08{)}^2=\pi \times 0.0064\approx 2.01\times {10}^{-2}{\text{m}}^2$$

Substitute values:

$$\tau =30\times 6.0\times (2.01\times {10}^{-2})\times 1.0\times \sin {60}^{\circ }$$
$$\sin {60}^{\circ }=\frac{\sqrt{3}}{2}\approx 0.866$$
$$\tau \approx 30\times 6.0\times 2.01\times {10}^{-2}\times 0.866$$
$$\tau \approx 3.1\text{ N m}$$

Answer (a)

$$\boxed{{\displaystyle \tau \approx 3.1\text{ N m}}}$$

This is the counter torque that must be applied to keep the coil from rotating.

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(b)

Would your answer change if the circular coil were replaced by a planar coil of some irregular shape enclosing the same area (all other particulars unchanged)?

Answer

No, the answer would not change.

Explanation

The torque on a planar current-carrying coil depends on:

$$\tau =NIAB\sin \theta$$

It depends only on:

• number of turns $N$,

• current $I$,

• enclosed area $A$,

• magnetic field $B$,

• angle $\theta$ between the field and the normal.

It is independent of the shape of the coil.
Hence, any planar coil (circular or irregular) enclosing the same area will experience the same torque under identical conditions.

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