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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 centre allowing it to turn in a horizontal plane. - Physics

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प्रश्न

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 centre allowing it to turn in a horizontal plane.

  1. What is the magnetic moment associated with the solenoid?
  2. 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?
संख्यात्मक
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उत्तर

Number of turns on the solenoid, n = 2000

Area of cross-section of the solenoid, A = 1.6 × 10−4 m2

Current in the solenoid, I = 4 A

  1. The magnetic moment along the axis of the solenoid is calculated as:
    M = nAI
    = 2000 × 1.6 × 10−4 × 4
    = 1.28 Am2
  2. Magnetic field, B = 7.5 × 10−2 T
    Angle between the magnetic field and the axis of the solenoid, θ = 30°
    Torque, τ = MB sin θ
    = 1.28 × 7.5 × 10−2 × sin 30°
    = 4.8 × 10−2 Nm
    Since the magnetic field is uniform, the force on the solenoid is zero.
    The torque on the solenoid is 4.8 × 10−2 Nm.
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पाठ 5: Magnetism and Matter - EXERCISES [पृष्ठ १५२]

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एनसीईआरटी Physics [English] Class 12
पाठ 5 Magnetism and Matter
EXERCISES | Q 5.6 | पृष्ठ १५२

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