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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 - Physics

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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?

योग
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उत्तर

Given:
No. of turns of the coil, n = 50
Magnetic field intensity, B = 0.20 T = 2 × 10−1 T
Radius of the coilr = 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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अध्याय 12: Magnetic Field - Exercises [पृष्ठ २३४]

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एचसी वर्मा Concepts of Physics Vol. 2 [English] Class 11 and 12
अध्याय 12 Magnetic Field
Exercises | Q 54 | पृष्ठ २३४
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