If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?
Magnetic field strength, B = 0.25 T
Magnetic moment, M= 0.6 T-1
The angle θ, between the axis of the solenoid and the direction of the applied field is 30°.
Therefore, the torque acting on the solenoid is given as:
T = MB sinθ
= 0.6 x 0.25 sin30°
= 7.5 x 102 J
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
Welcome to the NCERT Solutions for Class 12 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 6: If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizonta....
Comments