Question

# A 50-turn circular coil of radius 2.0 cm carrying a current of 5.0 A is rotated in a magnetic field of strength 0.20 T. (a) What is the maximum torque that acts on the coil? (b) In a particular position of the coil, the torque acting on it is half of this maximum. What is the angle between the magnetic field and the plane of the coil?

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Solution

## Given: No. of turns of the coil, n = 50 Magnetic field intensity, B = 0.20 T = 2 × 10−1 T Radius of the coil, r = 0.02 m = 2 × 10−2 m Magnitude of current =5 A Torque acting on the coil, τ = niABsinθ Here, A is the area of the coil and θ is the angle between the area vector and the magnetic field. τ is maximum when θ = 90°. τmax = niABsin90° = 50 × 5 × 3.14 × 4 × 10−4 × 2 × 10−1 = 6.28 × 10−2 N-m $\mathrm{Given},\tau =\frac{1}{2}×{\tau }_{max}\phantom{\rule{0ex}{0ex}}⇒\mathrm{sin\theta }=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒\theta \mathit{}=30°$ So, the angle between the magnetic field and the plane of the coil = 90° − 30° = 60°

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