JEE Main Magnetic Effect of Current PYQ

51. (JEE Main 2019 (Online) 10th April Evening Slot )

A square loop is carrying a steady current I and the magnitude of its magnetic dipole moment is m. if this square loop is changed to a circular loop and it carries the same current, the magnitude of the magnetic dipole moment of circular loop will be:

A. $\frac{m}{π}$

B. $\frac{3m}{π}$

C. $\frac{2m}{π}$

D. $\frac{4m}{π}$

Correct Answer is Option ()

52. (JEE Main 2019 (Online) 10th January Evening Slot )

A hoop and a solid cylinder of same mass and radius are made of a permanent magnetic material with their magnetic moment parallel to their respective axes. But the magnetic moment of hoop is twice of solid cylinder. They are placed in a uniform magnetic field in such a manner that their magnetic moments make a small angle with the field. If the oscillation periods of hoop and cylinder are T_h and T_c respectively, then -

A. T_h = 1.5 T_c

B. T_h = T_c

C. T_h = 2T_c

D. T_h = 0.5 T_c

Correct Answer is Option (B)

T = $2 π \sqrt{\frac{1}{μ B}}$

T_h = $2 π \sqrt{\frac{m R^{2}}{(2 μ) B}}$

T_C = $2 π \sqrt{\frac{1 / 2 m R^{2}}{μ B}}$

53. (JEE Main 2019 (Online) 10th January Morning Slot )

An insulating thin rod of length $l$ has a linear charge density $ρ (x)$ = $ρ_{0} \frac{x}{l}$ on it. The rod is rotated about an axis passing through the origin (x = 0) and perpendicular to the rod. If the rod makes n rotations per second, then the time averaged magnetic moment of the rod is -

A. $\frac{π}{3} n ρ l^{3}$

B. $\frac{π}{4} n ρ l^{3}$

C. $n ρ l^{3}$

D. $π n ρ l^{3}$

Correct Answer is Option (B)

$∵$ M = NIA

dq = $λ$ dx & A = $π$ x²

$\int d m = \int (x) \frac{ρ_{0} x}{ℓ} d x . π x^{2}$

M = $\frac{n ρ_{0} π}{ℓ} . \int_{0}^{ℓ} x^{3} . d x = \frac{n ρ_{0} π}{ℓ} . [\frac{L^{4}}{4}]$

M = $\frac{n ρ_{0} π ℓ^{3}}{4}$ or $\frac{π}{4} n ρ ℓ^{3}$

54. (JEE Main 2018 (Online) 16th April Morning Slot )

A charge q is spread uniformly over an insulated loop of radius r. If it is rotated with an angular velocity ω with resect to normal axis then the magnetic moment of the loop is :

A. q $ω$ r²

B. $\frac{4}{3}$ q $ω$ r²

C. $\frac{3}{2}$ q $ω$ r²

D. $\frac{1}{2}$ q $ω$ r²

Correct Answer is Option (D)

Magnetic moment,

$μ$ = I A

= $\frac{q}{T} (π r^{2})$

= $\frac{q}{2 π / ω} (π r^{2})$

= $\frac{q w}{2 π}$ $(π r^{2})$

= $\frac{1}{2}$ q $ω$ r²

55. (JEE Main 2018 (Offline) )

The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop is B₁. When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is $B_{2}$ . The ratio $\frac{B_{1}}{B_{2}}$ is:

A. 2

B. $\sqrt{3}$

C. $\sqrt{2}$

D. $\frac{1}{\sqrt{2}}$

Correct Answer is Option (C)

Dipole moment, M = IA

Let radius of circular loop = R

$∴$ M = I $\times$ $π$ R²

Later, we keep current constant ,

But dipole moment becomes double, let new radius = R₁

$∴$ 2M = I $\times$ $π$ R $_{1}^{2}$

$\Rightarrow$ 2I $π$ R² = I $π$ R $_{1}^{2}$

$\Rightarrow$ R₁ = $\sqrt{2}$ R

At the center of circular ring, the magnetic field,

B = $\frac{μ_{0} I}{2 R}$

$∴$ B₁ = $\frac{μ_{0} I}{2 R}$ and B₂ = $\frac{μ_{0} I}{2 \times (\sqrt{2} R)}$

$∴$ $\frac{B_{1}}{B_{2}}$ = $\frac{\frac{μ_{0} I}{2 R}}{\frac{μ_{0} I}{2 \sqrt{2} R}}$

= $\frac{2 \sqrt{2}}{2}$

= $\sqrt{2}$