A massless square loop of wire of resistance supporting a mass of 1 g hangs vertically with one of its sides in a uniform magnetic field of , directed outwards in the shaded region. [2023]

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Solution

Q.
A massless square loop of wire of resistance $10 Ω$ supporting a mass of 1 g hangs vertically with one of its sides in a uniform magnetic field of $10^{3} G$ , 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 [2023]

(If sides of the loop = 10 cm, $g = 10 {ms}^{- 2}$ )

1 $\frac{1}{10} V$

2 $100 V$

3 $1 V$

4 $10 V$

Ans.
(4)

$F_{m} = m g$

$∴$ $I L B = m g$

$∴$ $(\frac{v}{R}) L B = m g$

$∴$ $v = \frac{m g R}{L B} = \frac{(1 \times 10^{- 3} kg) (10 {m/s}^{2}) (10 Ω)}{(0.1 m) (10^{3} \times 10^{- 4} T)} = 10 V$

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