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Karnataka Board PUCPUC Science Class 11

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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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?

Sum

Solution

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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Chapter 12: Magnetic Field - Exercises [Page 234]

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HC Verma Concepts of Physics Vol. 2 [English] Class 11 and 12
Chapter 12 Magnetic Field
Exercises | Q 54 | Page 234

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