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Question 5.6: 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?Class 12 - Physics - Magnetism And Matter Magnetism And Matter Page-201

Answer» \text{as you know, solenoid behaves as a bar magnet}\\\text{so,torque on the solenoid},\vec{\tau}=\vec{M}\times\vec{<klux>B</klux>}\\|\tau|=MBsin\theta
A/C to QUESTION,
axis of solenoid makes an angle of 30° with the direction of APPLIED field. so, \theta=30^{\circ}
MAGNETIC field, B = 0.25T
from PREVIOUS question, M = 0.6 Am²
now, \tau=0.6\times0.25sin30{\circ}\\=0.075Nm


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