Solved: PHYSICS Question for NEET | Sushrut

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NEET PHYSICSMedium

A rectangular coil of length $0.12 \text{ m}$ and width $0.1 \text{ m}$ having $50$ turns of wire is suspended vertically in a uniform magnetic field of strength $0.2 \text{ Wb/m}^2$ . The coil carries a current of $2 \text{ A}$ . If the plane of the coil is inclined at an angle of $30^\circ$ with the direction of the field, the torque required to keep the coil in stable equilibrium will be:

A

0.15 N-m

B

0.20 N-m

C

0.24 N-m

D

0.12 N-m

Step-by-Step Solution

Formula: The torque ( $\tau$ ) exerted on a current-carrying coil in a magnetic field is given by $\tau = NIAB \sin \theta$ , where:

$N$ is the number of turns.
$I$ is the current.
$A$ is the area of the coil.
$B$ is the magnetic field strength.
$\theta$ is the angle between the normal to the coil's area and the magnetic field .

Identify Angle: The problem states the plane of the coil makes an angle of $30^\circ$ with the field. Therefore, the angle between the normal to the coil and the field is $\theta = 90^\circ - 30^\circ = 60^\circ$ .
Calculation:

Area $A = \text{length} \times \text{width} = 0.12 \times 0.1 = 0.012 \text{ m}^2$ .
Substituting values: $\tau = 50 \times 2 \times 0.012 \times 0.2 \times \sin(60^\circ)$ .
$\tau = 0.24 \times \frac{\sqrt{3}}{2} \approx 0.24 \times 0.866$ .
$\tau \approx 0.2078 \text{ N-m}$ .

Conclusion: The value closest to the calculated result is $0.20 \text{ N-m}$ .

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