JEE Main Magnetic Effect of Current PYQ

1. (JEE Main 2024 (Online) 1st February Evening Shift )

A moving coil galvanometer has 100 turns and each turn has an area of $2.0 {cm}^{2}$ . The magnetic field produced by the magnet is $0.01 T$ and the deflection in the coil is 0.05 radian when a current of $10 mA$ is passed through it. The torsional constant of the suspension wire is $x \times 10^{- 5} N - m / rad$ . The value of $x$ is _______ .

Correct answer is 4

$\begin{aligned} τ = BINAsin ϕ \\ C θ = BINAsin 90^{\circ} \\ C = \frac{BINA}{θ} = \frac{0.01 \times 10 \times 10^{- 3} \times 100 \times 2 \times 10^{- 4}}{0.05} \\ = 4 \times 10^{- 5} N - m / rad . \\ x = 4 \end{aligned}$

2. (JEE Main 2023 (Online) 25th January Evening Shift )

For a moving coil galvanometer, the deflection in the coil is 0.05 rad when a current of 10 mA is passes through it. If the torsional constant of suspension wire is $4.0 \times 10^{- 5} N m {rad}^{- 1}$ , the magnetic field is 0.01T and the number of turns in the coil is 200, the area of each turn (in cm $^{2}$ ) is :

A. 1.5

B. 2.0

C. .05

D. 1.0

Correct Answer is Option (D)

$∵ θ = (\frac{N B A}{K}) I$

$\begin{aligned} A & = \frac{θ K}{N B I} \\ = \frac{0.05 \times 4 \times 10^{- 5}}{(200) \times (0.01) \times (10 \times 10^{- 3})} \\ = 1 {cm}^{2} \end{aligned}$

3. (JEE Main 2022 (Online) 28th July Morning Shift )

The current sensitivity of a galvanometer can be increased by :

(A) decreasing the number of turns

(B) increasing the magnetic field

(C) decreasing the area of the coil

(D) decreasing the torsional constant of the spring

Choose the most appropriate answer from the options given below :

A. (B) and (C) only

B. (C) and (D) only

C. (A) and (C) only

D. (B) and (D) only

Correct Answer is Option (D)

$N i A B = k θ$

$\Rightarrow \frac{θ}{i} = \frac{N A B}{k}$

$\Rightarrow$ Sensitivity increases if $B ↑$ and $k ↓$

4.(JEE Main 2020 (Online) 3rd September Evening Slot)

A galvanometer coil has 500 turns and each turn has an average area of 3 $\times$ 10^–4 m² . If a torque of 1.5 Nm is required to keep this coil parallel to a magnetic field when a current of 0.5 A is flowing through it, the strength of the field (in T) is ______.

Correct Answer is 20

Given N = 500

A = 3 $\times$ 10^–4 m²

$τ$ = 1.5 Nm

i = 0.5 A

We know, $τ = B I N A s i n θ$

$1.5 = B \times 0.5 \times 500 \times 3 \times 10^{- 4}$

$B = \frac{10000}{500} = 20$ Tesla

5. (JEE Main 2019 (Online) 9th April Evening Slot )

A moving coil galvanometer has a coil with 175 turns and area 1 cm². It uses a torsion band of torsion constant 10^–6 N-m/rad. The coil is placed in a maganetic field B parallel to its plane. The coil deflects by 1° for a current of 1 mA. The value of B (in Tesla) is approximately :-

A. 10^–4

B. 10^–2

C. 10^–1

D. 10^–3

Correct Answer is Option (D)

NIAB = KQ

175 × 1 × 10^–3 × 1 × 10^–4 × B = $\frac{10^{- 6} \times π}{180}$

$\Rightarrow B = \frac{π}{180} \times \frac{10}{175} \approx 9.97 \times 10^{- 4} T$

$\Rightarrow$ B = 10^–3 T