5.8 A closely wound solenoid of 2000 turns and area of cross-section 1.6 \times 10 ^{-4} m ^2, carrying a current of 4.0 A, is suspended through its
centre allowing it to turn in a horizontal plane.  

(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 \times 10 ^{-2} T  is set up at an angle of 30° with the axis of the solenoid?

Answers (1)

Now,

Magnetic field strength, B = 7.5 \times 10 ^{-2} T

The angle between the magnetic field and the axis of the solenoid, \theta = 30^{\circ}

Now, As the Magnetic field is uniform, the Force is zero

Also, we know,

\tau = mxB = mBsinθ

= (1.28 JT^{-1})(7.5 \times 10 ^{-2} T)(sin 30^{\circ})

= 4.8 x 10^{-2} J

Therefore, Force on the solenoid = 0 and torque on the solenoid = 4.8 x 10^{-2} J

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