Q. 544.0( 2 Votes )

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?

Answer :


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



We know that torque acting on a rectangular coil having n turns is given by



Where


B= applied magnetic field


A= area of rectangular loop


I = current flowing through coil


= angle between the area vector and magnetic field


Torque is maximum if


θ = 90°.





Given that ,





So, the angle between the magnetic field and the plane of coil is given by -



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