Ncert Exercises & Solutions

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?

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:

\begin{aligned} τ & = M B \sin θ \\ = 0.6 \times 0.25 \sin 30^{\circ} \\ = 7.5 \times 10^{- 2} J \end{aligned}

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