Magnetic Effect of Current Questions — NEET Physics

Q1

A 100-turn closely wound circular coil of radius 5 cm has a magnetic field of

3.14 \times 10^{-3} \mathrm{~T}

at its centre. The current flowing through the coil, and the magnitude of the magnetic moment of this coil are, respectively : (Take

\mu_0=4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~m} / \mathrm{A}

)

A

2 \mathrm{~A}, 10 \mathrm{~A} \mathrm{~m}^2

B

2.5 \mathrm{~A}, 20 \mathrm{~A} \mathrm{~m}^2

C

2 \mathrm{~A}, 4 \mathrm{~A} \mathrm{~m}^2

D

2.5 \mathrm{~A}, 2 \mathrm{~A} \mathrm{~m}^2

Correct Answer

Option D

Solution

Magnetic field of a circular loop:

\begin{aligned} & B_0=\frac{\mu_0 N i}{2 R} \\ & i=\frac{2 R B_0}{\mu_0 N}=\frac{2 \times 5 \times 10^{-2} \times 3.14 \times 10^{-3}}{4 \pi \times 10^{-7} \times 100} \\ & i=2.5 \mathrm{~A} \end{aligned}

Magnetic moment

\begin{aligned} & M=N i A \\ & =100 \times 2.5 \times 3.14 \times\left(5 \times 10^{-2}\right)^2\left[\because A=\pi R^2\right] \\ & =2 \mathrm{~A} \mathrm{~m}^2 \end{aligned}

Q2

A parallel plate capacitor made of circular plates is being charged such that the surface charge density on its plates is increasing at a constant rate with time. The magnetic field arising due to displacement current is:

A Non-zero everywhere with maximum at the imaginary cylindrical surface connecting peripheries of the plates

B Zero between the plates and non-zero outside

C Zero at all places

D Constant between the plates and zero outside the plates

Correct Answer

Option A

Solution

Let the surface charge density be $\sigma=\dfrac{q}{A}$ Given $\dfrac{d \sigma}{d t}=$ constant

\therefore \quad \frac{d}{d t}\left(\frac{q}{A}\right)=\text{ constant } \Rightarrow \frac{l}{A} i=\text{ constant }

It means displacement current is constant.

This system will act like a cylindrical wire.

The graph of magnetic field (B) vs $r$ is

Q3

A model for quantized motion of an electron in a uniform magnetic field

B

states that the flux passing through the orbit of the electron in

n(h l e)

where

n

is an integer,

h

is Planck's constant and

e

is the magnitude of electron's charge. According to the model, the magnetic moment of an electron in its lowest energy state will be (

m

is the mass of the electron)

A

\dfrac{h e B}{\pi m}

B

\dfrac{h e B}{2 \pi m}

C

\dfrac{h e}{\pi m}

D

\dfrac{h e}{2 \pi m}

Correct Answer

Option D

Solution

To understand the quantized motion of an electron in a uniform magnetic field and determine its magnetic moment in the lowest energy state, consider the following derivations and equations: Magnetic Force and Centripetal Force: The magnetic force acting on the electron is counterbalanced by the centripetal force necessary for its circular motion: $e v B = \dfrac{m v^2}{r}$ Solving for the velocity $v$ : $v = \dfrac{e B r}{m}$ Flux through the Electron's Orbit: The model states that the magnetic flux $\phi$ through the orbit is linked to Planck’s constant $h$ as: $B \pi r^2 = \dfrac{n h}{e}$ Rearranging gives: $B r^2 = \dfrac{n h}{e \pi}$ Magnetic Moment: The magnetic moment $\mu$ is defined as the current $I$ times the area $A$ of the orbit: $\mu = I A = \dfrac{e}{T} \pi r^2$ Considering current due to orbital motion: $\mu = \dfrac{e v}{2 \pi r} \times \pi r^2 = \dfrac{e v r}{2}$ Substitute for $v$ from the earlier velocity equation: $\mu = \dfrac{1}{2} e \left(\dfrac{e B r}{m}\right) r = \dfrac{1}{2} e^2 \dfrac{B r^2}{m}$ Using the flux condition $B r^2 = \dfrac{n h}{e \pi}$ , we find: $\mu = \dfrac{1}{2} e^2 \dfrac{n h}{e \pi m} = \dfrac{n e h}{2 \pi m}$ Lowest Energy State: For the lowest energy state, take $n = 1$ : $\mu = \dfrac{e h}{2 \pi m}$ Thus, the magnetic moment of an electron in its lowest energy state is $\dfrac{e h}{2 \pi m}$ .

Q4

An electron (mass

9 \times 10^{-31} \mathrm{~kg}

and charge

1.6 \times 10^{-19} \mathrm{C}

) moving with speed

c / 100(c=

speed of light) is injected into a magnetic field

\vec{B}

of magnitude

9 \times 10^{-4} \mathrm{~T}

perpendicular to its direction of motion. We wish to apply an uniform electric field

\vec{E}

together with the magnetic field so that the electron does not deflect from its path. Then (speed of light

c=3

\times 10^3 \mathrm{~ms}^{-1}

)

A

\vec{E}

is parallel to

\vec{B}

and its magnitude is

27 \times 10^2 \mathrm{~V} \mathrm{~m}^{-1}

B

\vec{E}

is parallel to

\vec{B}

and its magnitude is

27 \times 10^4 \mathrm{~V} \mathrm{~m}^{-1}

C

\vec{E}

is perpendicular to

\vec{B}

and its magnitude is

27 \times 10^4 \mathrm{~V} \mathrm{~m}^{-1}

D

\vec{E}

is perpendicular to

\vec{B}

and its magnitude is

27 \times 10^2 \mathrm{~V} \mathrm{~m}^{-1}

Correct Answer

Option D

Solution

For no deflection of electron, $\vec{F}_B=\vec{F}_E$

\begin{aligned} & -e(\vec{v} \times \vec{B})=-e \vec{E} \\ & \Rightarrow \vec{E}=\vec{v} \times \vec{B} \Rightarrow \vec{E} \perp \vec{B} \\ & E=v B=\frac{c}{100} \times 9 \times 10^{-4} \\ & =\frac{3 \times 10^8}{100} \times 9 \times 10^{-4} \\ & =27 \times 10^2 \mathrm{~V} \mathrm{~m}^{-1} \end{aligned}

Q5

A 2 amp current is flowing through two different small circular copper coils having radii ratio

1: 2

. The ratio of their respective magnetic moments will be

A

2: 1

B

4: 1

C

1: 4

D

1: 2

Correct Answer

Option C

Solution

The magnetic moment of a current-carrying circular loop is given by the formula $M = I \times A$ , where $I$ is the current and $A$ is the area of the loop.

Assuming the current ( $I$ ) is the same for both coils, the magnetic moment is directly proportional to the area of the coil ( $M \propto A$ ).

For two coils with radii $r_1$ and $r_2$ , the areas are $A_1 = \pi r_1^2$ and $A_2 = \pi r_2^2$ .

Given that the ratio of the radii is $1:2$ , we can write: $\dfrac{M_1}{M_2} = \dfrac{A_1}{A_2} = \dfrac{\pi r_1^2}{\pi r_2^2} = \left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4}$ Thus, the ratio of their respective magnetic moments is $1:4$ .

Q6

A tightly wound 100 turns coil of radius

10 \mathrm{~cm}

carries a current of

7 \mathrm{~A}

. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as

4 \pi \times 10^{-7} \mathrm{SI}

units):

A 44 mT

B 4.4 T

C 4.4 mT

D 44 T

Correct Answer

Option C

Solution

The magnitude of magnetic field due to circular coil of N turns is given by

\begin{aligned} & B_C=\frac{\mu_0 i N}{2 R} \\ & =\frac{4 \pi \times 10^{-7} \times 7 \times 100}{2 \times 0.1} \\ & =4.4 \times 10^{-3} \mathrm{~T} \\ & =4.4 \mathrm{~mT} \end{aligned}

Q7

A parallel plate capacitor is charged by connecting it to a battery through a resistor. If

I

is the current in the circuit, then in the gap between the plates:

A There is no current

B Displacement current of magnitude equal to

I

flows in the same direction as

I

C Displacement current of magnitude equal to

I

flows in a direction opposite to that of

I

D Displacement current of magnitude greater than

I

flows but can be in any direction

Correct Answer

Option B

Solution

According to modified Ampere's law

\oint B \cdot d I=\mu_0\left(I_C+I_D\right)

For Loop

L_1 \quad I_C \neq 0

and

I_D=0

For Loop

L_2 \quad I_C=0

and

I_D \neq 0

Due to

\mathrm{KCL} \quad I_C=I_D

Q8

A long straight wire of length

2 \mathrm{~m}

and mass

250 \mathrm{~g}

is suspended horizontally in a uniform horizontal magnetic field of

0.7 \mathrm{~T}

. The amount of current flowing through the wire will be

\left(\mathrm{g}=9.8 \mathrm{~ms}^{-2}\right)

:

A 2.45 A

B 2.25 A

C 2.75 A

D 1.75 A

Correct Answer

Option D

Solution

\begin{aligned} & \mathrm{m g=i \ell B} \\\\ & 250 \times 10^{-3} \times 9.8=\mathrm{i} \times 2 \times 0.7 \\\\ & \mathrm{i}=\frac{0.250 \times 9.8}{2 \times 0.7}=0.250 \times 7 \\\\ & \mathrm{i}=1.75 \mathrm{~A} \end{aligned}

Q9

A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron:

A will turn towards right of direction of motion

B will turn towards left of direction of motion

C speed will decrease

D speed will increase

Correct Answer

Option C

Solution

Speed of electron will decrease due to electric force magnetic force of electron is zero.

Q10

A wire carrying a current

I

along the positive

\mathrm{x}

-axis has length

L

. It is kept in a magnetic field

\overrightarrow{\mathrm{B}}=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}) \mathrm{T}

. The magnitude of the magnetic force acting on the wire is :

A

\sqrt{5} \mathrm{~IL}

B

5 \mathrm{~IL}

C

\sqrt{3}

IL

D 3 IL

Correct Answer

Option B

Solution

\vec{F}=I(\vec{\ell} \times \vec{B})

\begin{aligned} & =I[(L \hat{i}) \times(2 \hat{i}+3 \hat{j}-4 \hat{k})] \\\\ & =I(4 L \hat{j}+3 L \hat{k}) \end{aligned}

|\vec{F}|=5 \text{ IL }