JEE Main Magnetism and Matter PYQ

1. (JEE Main 2024 (Online) 6th April Evening Shift)

Match List I with List II:

	LIST I (Y vs X)
A.	Y-magnetic susceptibility X = magnetising field	I.
B.	Y = magnetic field X = distance from centre of a current carrying wire for x < a (where a = radius of wire)	II.
C.	Y = magnetic field $X =$ distance from centre of a current carrying wire for $x > a$ (where $a =$ radius of wire)	III.
D.	Y = magnetic field inside solenoid X = distance from centre	IV.

Choose the correct answer from the options given below :

A.(A)-(III), (B)-(IV), (C)- (I), (D)-(II)

B.(A)-(III), (B)-(I), (C)-(IV), (D)-(II)

C.(A)-(IV), (B)-(I), (C)-(III), (D)-(II)

D.(A)-(I), (B)-(III), (C)-(II), (D)-(IV)

Correct option is (B)

In para and dia materials, susceptibility is independent of magnetic field.

Inside a current carrying wire, $B \propto X$

Outside a current carrying wire, $B \propto \frac{1}{x}$

Inside a solenoid, B is constant upto ends.

2. (JEE Main 2024 (Online) 8th April Evening Shift )

The coercivity of a magnet is $5 \times 10^{3} A / m$ . The amount of current required to be passed in a solenoid of length $30 cm$ and the number of turns 150, so that the magnet gets demagnetised when inside the solenoid is ________ A.

Correct answer is 10

Coercivity is a measure of the resistance of a ferromagnetic material to becoming demagnetized. It is defined as the intensity of the applied magnetic field required to reduce the magnetization of a material to zero after the magnetization of the sample has been driven to saturation. In this case, coercivity $H_{c}$ is given to be $5 \times 10^{3} A / m$ .

The relationship between the magnetic field $H$ inside a solenoid and the current $I$ passed through it is given by the formula:

$H = \frac{N \cdot I}{L}$

where:

$H$ is the magnetic field strength inside the solenoid in amperes per meter (A/m),
$N$ is the total number of turns of wire,
$I$ is the current in amperes (A), and
$L$ is the length of the solenoid in meters (m).

Given that the number of turns of the solenoid $N = 150$ and the length of the solenoid $L = 30 cm = 0.3 m$ , we can rearrange the formula to solve for $I$ :

$I = \frac{H \cdot L}{N}$

Substituting the given values:

$I = \frac{(5 \times 10^{3}) \times 0.3}{150}$

Simplifying, we get:

$I = \frac{1500}{150}$

$I = 10 A$

Therefore, the amount of current required to be passed in the solenoid for demagnetizing the magnet when inside the solenoid is 10 A.

3. (JEE Main 2023 (Online) 11th April Morning Shift)

The free space inside a current carrying toroid is filled with a material of susceptibility $2 \times 10^{- 2}$ . The percentage increase in the value of magnetic field inside the toroid will be

A.0.1%

B.1%

C.2%

D.0.2%

Correct option is (C)

The magnetic susceptibility ( $χ_{m}$ ) of a material is the measure of how much the material becomes magnetized in response to an external magnetic field. When the material is placed in a magnetic field, the net magnetic field ( $B$ ) inside the material is the sum of the external magnetic field ( $B_{0}$ ) and the field produced by the material itself ( $B_{m a t e r i a l}$ ). This relationship can be expressed as :

$B = B_{0} + B_{m a t e r i a l}$

Magnetic susceptibility is related to the relative permeability ( $μ_{r}$ ) of the material :

$μ_{r} = 1 + χ_{m}$

The magnetic field inside the toroid can be calculated using the following formula :

$B = μ_{0} μ_{r} H$

where $μ_{0}$ is the permeability of free space, $μ_{r}$ is the relative permeability of the material, and $H$ is the magnetic field strength.

Now, let's consider the percentage increase in the magnetic field when the material is placed inside the toroid. The initial magnetic field ( $B_{0}$ ) is given by :

$B_{0} = μ_{0} H$

After the material is placed inside the toroid, the magnetic field becomes :

$B = μ_{0} μ_{r} H = μ_{0} (1 + χ_{m}) H$

The percentage increase in the magnetic field can be calculated as :

$\frac{B - B_{0}}{B_{0}} \times 100 % = \frac{μ_{0} (1 + χ_{m}) H - μ_{0} H}{μ_{0} H} \times 100 %$

Substitute the value of $χ_{m} = 2 \times 10^{- 2}$ :

$\frac{B - B_{0}}{B_{0}} \times 100 % = \frac{(1 + 2 \times 10^{- 2}) - 1}{1} \times 100 % = 2 %$

Thus, the percentage increase in the value of the magnetic field inside the toroid is 2%.

4. (JEE Main 2023 (Online) 25th January Morning Shift)

A solenoid of 1200 turns is wound uniformly in a single layer on a glass tube 2 m long and 0.2 m in diameter. The magnetic intensity at the center of the solenoid when a current of 2 A flows through it is :

A. 1 A m^-1

B. $2.4 \times 10^{- 3} A m^{- 1}$

C.

1.2 \times 10^{3} A m^{- 1}

D. $2.4 \times 10^{3} A m^{- 1}$

Correct Option is (C)

Number of turns per unit length $= \frac{1200}{2} = 600$

So, Magnetic Intensity $H = n I$

$\begin{aligned} = 600 \times 2 {Am}^{- 1} \\ = 1200 {Am}^{- 1} \end{aligned}$

5. (JEE Main 2023 (Online) 10th April Morning Shift )

The current required to be passed through a solenoid of 15 cm length and 60 turns in order of demagnetise a bar magnet of magnetic intensity $2.4 \times 10^{3} A m^{- 1}$ is ___________ A.

Correct answer is (6)

We know that the magnetizing field (H) inside a solenoid is given by the formula :

$H = \frac{N \cdot I}{L}$

where (N) is the number of turns, (I) is the current in Amperes, and (L) is the length of the solenoid in meters.

To demagnetize a bar magnet that has a magnetic intensity (H) of ( $2.4 \times 10^{3} {Am}^{- 1}$ ), you need to create a magnetizing field in the solenoid that is equal in magnitude but opposite in direction.

Given:

( $H = 2.4 \times 10^{3} {Am}^{- 1}$ )
(N = 60) turns
( $L = 15 cm = 0.15 m$ ) (since $(1 m = 100 cm)$ )

You can rearrange the formula for (H) to solve for (I) :

$I = \frac{H \cdot L}{N}$

Substituting the given values, you get :

$I = \frac{(2.4 \times 10^{3} {Am}^{- 1}) \cdot 0.15 m}{60}$

$I = 6 Amperes$

So, the current required to be passed through the solenoid to demagnetize the bar magnet is (6 $A$ ).