JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Magnetic Effects of Current

A massless square loop, of wire of resistance

10Ω

. supporting a mass of I g. hangs vertically with one of its sides in a uniform magnetic field of

1 0^{3} G

, directed outwards in the shaded region. A dc voltage V is applied to the loop. For what value of V . the magnetic force will exactly balance the weight of the supporting mass of 1 g ?
(If sides of the loop

= 10 cm, g = 10 ms^{- 2}

)

(A) $\frac{1}{10} V$
(B)100 V
(C)1 V
(D)10 V✓ Correct

Answer: (D)

Solution

For the loop to hang in balance the magnetic force on the current-carrying side lying in the field must support the weight:

B I L = m g

. Writing the current as

I = \frac{V}{R}

, this gives

V = \frac{m g R}{B L}

. With

m = 1 0^{- 3} kg

g = 10 m/s^{2}

R = 10 Ω

L = 0.1 m

and

B = 1 0^{3} G = 0.1 T

V = \frac{1 0 ^{- 3} \times 10 \times 10}{0.1 \times 0.1} = 10 V

. The required voltage is

10 V

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Single correct option

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JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Magnetic Effects of Current

Solution

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