Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A 250-turn rectangular coil with a length of $2.1 \text{ cm}$ and a width of $1.25 \text{ cm}$ carries a current of $85 \text{ }\mu\text{A}$ and is subjected to a magnetic field of strength $0.85 \text{ T}$ . What is the work done to rotate the coil by $180^\circ$ against the torque?

9.1 $\mu$ J

4.55 $\mu$ J

2.3 $\mu$ J

1.5 $\mu$ J

Step-by-Step Solution

Magnetic Moment ( $m$ ): The current-carrying coil behaves as a magnetic dipole. The magnetic moment is given by $m = NIA$ , where $N$ is the number of turns, $I$ is the current, and $A$ is the area of the coil.

$N = 250$
$I = 85 \text{ }\mu\text{A} = 85 \times 10^{-6} \text{ A}$
$A = \text{length} \times \text{width} = (2.1 \times 10^{-2} \text{ m}) \times (1.25 \times 10^{-2} \text{ m}) \approx 2.625 \times 10^{-4} \text{ m}^2$ $m = 250 \times (85 \times 10^{-6}) \times (2.625 \times 10^{-4}) \approx 5.58 \times 10^{-6} \text{ A m}^2$

Work Done ( $W$ ): The work done to rotate a magnetic dipole in a uniform magnetic field $B$ from an initial angle $\theta_1$ to a final angle $\theta_2$ is equal to the change in potential energy ( $U = -mB \cos \theta$ ). Rotating 'against the torque' typically implies moving from the stable equilibrium ( $\theta_1 = 0^\circ$ ) to the unstable equilibrium ( $\theta_2 = 180^\circ$ ). $W = U_f - U_i = -mB(\cos 180^\circ - \cos 0^\circ)$ $W = -mB(-1 - 1) = 2mB$
Calculation: $W = 2 \times (5.58 \times 10^{-6}) \times 0.85$ $W \approx 9.48 \times 10^{-6} \text{ J} = 9.48 \text{ }\mu\text{J}$ The calculated value ( $9.48 \text{ }\mu\text{J}$ ) is closest to the option $9.1 \text{ }\mu\text{J}$ . (Discrepancy likely due to rounding of area dimensions in the original question source).

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