# 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?

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 $\dpi{100} JT^{-1}$)($\dpi{100} 7.5 \times 10 ^{-2} T$)(sin $30^{\circ}$)

= 4.8 x $\dpi{100} 10^{-2}$ J

Therefore, Force on the solenoid = 0 and torque on the solenoid = 4.8 x $\dpi{100} 10^{-2}$ J

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