Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

A 250 turn rectangular coil of length $2.1 \text{ cm}$ and width $1.25 \text{ cm}$ carries a current of $85 \, \mu\text{A}$ and subjected to the magnetic field of strength $0.85 \text{ T}$ . Work done for rotating the coil by $180^{\circ}$ against the torque is:

$4.55 \, \mu\text{J}$

$2.3 \, \mu\text{J}$

$1.15 \, \mu\text{J}$

$9.4 \, \mu\text{J}$

Step-by-Step Solution

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

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

Work Done ( $W$ ): The work done in rotating a magnetic dipole from an initial angle $\theta_1$ to a final angle $\theta_2$ in a uniform magnetic field $B$ is given by the change in potential energy: $W = \Delta U = mB(\cos\theta_1 - \cos\theta_2)$ .

"Rotating by $180^{\circ}$ against the torque" typically implies rotating from the stable equilibrium position ( $\theta_1 = 0^{\circ}$ ) to the unstable equilibrium position ( $\theta_2 = 180^{\circ}$ ).
$W = mB(\cos 0^{\circ} - \cos 180^{\circ}) = mB(1 - (-1)) = 2mB$

Calculation:

$W = 2 \times (5.58 \times 10^{-6}) \times 0.85$
$W \approx 9.48 \times 10^{-6} \text{ J} = 9.48 \, \mu\text{J}$

Conclusion: The value closest to the calculated work is $9.4 \, \mu\text{J}$ .

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