Ncert Exercises & Solutions

5.8 A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10–4 m2, carrying a current of 4.0 A, is suspended through its center allowing it to turn in a horizontal plane.

(a) What is the magnetic moment associated with the solenoid?

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

Number of turn on the solenoid, n=2000

Area of cross-section of the solenoid,

A = 1.6 \times 10^{- 4} m^{2}

Current In the solenoid, I = 4A

(a) The magnetic field moment along the axis of the solenoid is calculated as:

M = n A I = 2000 \times 1.6 \times 10^{- 4} \times 4 = 1.28 {Am}^{2}

(b) Magnetic field,

B = 7.5 \times 10^{- 2} T

Angle between the magnetic field and the axis of the solenoid,

θ = 30 °

Torque,

τ = MB \sin θ

\begin{array}{l} = 1.28 \times 7.5 \times 10^{- 2} \sin 30^{\circ} \\ = 4.8 \times 10^{- 2} Nm \end{array}

Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is

4.8 \times 10^{- 2} Nm .

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