JEE Main Magnetism and Matter PYQ

1. (JEE Main 2024 (Online) 4th April Evening Shift)

The magnetic moment of a bar magnet is $0.5 {Am}^{2}$ . It is suspended in a uniform magnetic field of $8 \times 10^{- 2} T$ . The work done in rotating it from its most stable to most unstable position is:

A. $4 \times 10^{- 2} J$

B. $16 \times 10^{- 2} J$

C. $8 \times 10^{- 2} J$

D.Zero

Correct option is (C)

$\begin{aligned} W & = \int_{0}^{180} d τ \cdot d θ \\ = m B (\cos 0 - \cos 180) \\ = 0.5 \times 8 \times 10^{- 2} (2) \\ = 8 \times 10^{- 2} J \end{aligned}$

2. (JEE Main 2024 (Online) 30th January Morning Shift )

The horizontal component of earth's magnetic field at a place is $3.5 \times 10^{- 5} T$ . A very long straight conductor carrying current of $\sqrt{2} A$ in the direction from South east to North West is placed. The force per unit length experienced by the conductor is __________ $\times 10^{- 6} N / m$ .

Correct answer is 35

$\begin{aligned} B_{H} = 3.5 \times 10^{- 5} T \\ F = i ℓ B \sin θ, i = \sqrt{2} A \\ \frac{F}{ℓ} = i B \sin θ = \sqrt{2} \times 3.5 \times 10^{- 5} \times \frac{1}{\sqrt{2}} \\ = 35 \times 10^{- 6} N / m \end{aligned}$

3. (JEE Main 2024 (Online) 29th January Morning Shift )

The magnetic potential due to a magnetic dipole at a point on its axis situated at a distance of $20 cm$ from its center is $1.5 \times 10^{- 5} T m$ . The magnetic moment of the dipole is _________ $A m^{2}$ . (Given : $\frac{μ_{o}}{4 π} = 10^{- 7} Tm A^{- 1}$ )

Correct answer is 6

$\begin{aligned} V = \frac{μ_{0}}{4 π} \frac{M}{r^{2}} \\ \Rightarrow 1.5 \times 10^{- 5} = 10^{- 7} \times \frac{M}{{(20 \times 10^{- 2})}^{2}} \\ \Rightarrow M = \frac{1.5 \times 10^{- 5} \times 20 \times 20 \times 10^{- 4}}{10^{- 7}} \\ M = 1.5 \times 4 = 6 \end{aligned}$

4. (JEE Main 2023 (Online) 10th April Evening Shift)

A bar magnet is released from rest along the axis of a very long vertical copper tube. After some time the magnet will

A.move down with almost constant speed

B.move down with an acceleration equal to g

C.move down with an acceleration greater than g

D.oscillate inside the tube

Correct option is (A)

When a bar magnet is released from rest along the axis of a very long vertical copper tube, it will move down with an almost constant speed after some time.

As the magnet falls, it moves through the copper tube, inducing eddy currents in the tube. These eddy currents, in turn, create an opposing magnetic field that opposes the motion of the magnet. The opposing force generated by the eddy currents acts as a damping force, which slows down the magnet's acceleration. Eventually, the magnet reaches a terminal velocity, at which point the gravitational force pulling the magnet downward is balanced by the opposing force from the eddy currents.

At this terminal velocity, the magnet moves down with an almost constant speed.

5. (JEE Main 2023 (Online) 31st January Morning Shift)

A bar magnet with a magnetic moment $5.0 {Am}^{2}$ is placed in parallel position relative to a magnetic field of $0.4 T$ . The amount of required work done in turning the magnet from parallel to antiparallel position relative to the field direction is _____________.

A. zero

B. 1J

C. 2J

D. 4J

Correct Option is (D)

$W = - M B (\cos θ_{2} - \cos θ_{1})$

$\begin{aligned} = & - 0.4 \times 5 [\cos 180^{\circ} - \cos 0] \\ = & 4 J \end{aligned}$