JEE Main Magnetism and Matter PYQ

6. (JEE Main 2022 (Online) 27th July Morning Shift)

Two bar magnets oscillate in a horizontal plane in earth's magnetic field with time periods of $3 s$ and $4 s$ respectively. If their moments of inertia are in the ratio of $3 : 2$ , then the ratio of their magnetic moments will be:

A. 2 : 1

B. 8 : 3

C. 1 : 3

D. 27 : 16

Correct Option is (B)

$T = 2 π \sqrt{\frac{I}{M B_{H}}}$

$\Rightarrow \frac{T_{1}}{T_{2}} = \sqrt{\frac{I_{1}}{I_{2}}} \sqrt{\frac{M_{2}}{M_{1}}}$

$\Rightarrow \frac{3}{4} = \sqrt{\frac{3}{2}} \sqrt{\frac{M_{2}}{M_{1}}}$

$\Rightarrow \frac{M_{1}}{M_{2}} = \frac{3}{2} \times \frac{16}{9} = \frac{8}{3}$

7. (JEE Main 2022 (Online) 25th July Evening Shift)

An electron with energy 0.1 keV moves at right angle to the earth's magnetic field of 1 $\times$ 10^{$-$ 4} Wbm^{$-$ 2}. The frequency of revolution of the electron will be :

(Take mass of electron = 9.0 $\times$ 10^{$-$ 31} kg)

A. $1.6 \times 10^{5} Hz$

B. $5.6 \times 10^{5} Hz$

C. $2.8 \times 10^{6} Hz$

D. $1.8 \times 10^{6} Hz$

Correct Option is (C)

$T = \frac{2 π m}{B q}$

$\Rightarrow$ Frequency $f = \frac{B q}{2 π m}$

$= \frac{10^{- 4} \times 1.6 \times 10^{- 19}}{2 π \times 9 \times 10^{- 31}}$

$≃ 2.8 \times 10^{6}$ Hz

8. (JEE Main 2022 (Online) 26th June Evening Shift)

A bar magnet having a magnetic moment of 2.0 $\times$ 10⁵ JT^{$-$ 1}, is placed along the direction of uniform magnetic field of magnitude B = 14 $\times$ 10^{$-$ 5} T. The work done in rotating the magnet slowly through 60 $^{\circ}$ from the direction of field is :

A. 14 J

B. 8.4 J

C. 4 J

D. 1.4 J

Correct Option is (A)

$U = - \vec{M} . \vec{B}$

So $U_{f} - U_{i} = - M B (1 - \cos θ)$

$= - 14 J$

So $W = - Δ U = 14 J$

9. (JEE Main 2021 (Online) 27th July Morning Shift )

In a uniform magnetic field, the magnetic needle has a magnetic moment 9.85 $\times$ 10^{$-$ 2} A/m² and moment of inertia 5 $\times$ 10^{$-$ 6} kg m². If it performs 10 complete oscillations in 5 seconds then the magnitude of the magnetic field is _______________ mT. [Take $π$ ² as 9.85]

Correct answer is (8)

$T = 2 π \sqrt{\frac{I}{M B}}$

B = 80 $\times$ 10^{$-$ 4} = 8 mT

10. (JEE Main 2020 (Online) 4th September Morning Slot)

A small bar magnet placed with its axis at 30^o with an external field of 0.06 T experiences a torque of 0.018 Nm. The minimum work required to rotate it from its stable to unstable equilibrium position is :

A. 6.4 $\times$ 10^-2 J

B. 9.2 $\times$ 10^-3 J

C. 7.2 $\times$ 10^-2 J

D. 11.7 $\times$ 10^-3 J

Correct Option is (C)

Torque on a bar magnet :

$τ = M B \sin θ$

Here, $θ$ = 30º, I = 0.018 N-m, B = 0.06 T

$0.018 = M \times 0.06 \times 0.5$

$\Rightarrow M = 0.6 A m^{2}$

$W = U_{f} - U_{i}$

$= M B (\cos θ_{i} - \cos θ_{f})$

$= 0.6 \times 0.06 (1 - (- 1))$

$= 7.2 \times 10^{- 2} J$