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प्रश्न
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?
उत्तर
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
Given , ` tau = 1/2 xx tau_max`
⇒ `sintheta = 1/2`
⇒ θ = 30°
So, the angle between the magnetic field and the plane of the coil = 90° − 30° = 60°
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