Question
Download Solution PDFA circular coil of radius 0.50 m and 100 turns, carrying a current of 80 mA, is placed such that the normal to its plane makes an angle of 30° with a uniform magnetic field of 4.0 T. The magnitude of torque acting on the coil is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Torque:
- It is defined as the product of the magnitude of the force and the perpendicular distance of the line of action of a force from the axis of rotation.
- Formula, τ = rFsinθ , where r = perpendicular distance, F = force, θ = angle between force and perpendicular distance
- The SI unit of torque is N-m.
Torque in a current loop:
- Let us consider a rectangular loop such that it carries a current of magnitude I.
- If we place this loop in a magnetic field, it experiences a torque but no net force, quite similar to what an electric dipole experiences in a uniform electric field.
- The torque acting on the coil is calculated as, τ = NIAB sinθ where N = number of turns, I = current, A = area of the loop, B = magnetic field, θ = angle between area and magnetic field.
Calculation:
Given,
The number of turns, N = 100 turns
The current in the coil, I = 80 mA
Angle, θ = 30º
The radius of the circular coil, r = 0.50 m
The magnitude of the magnetic field, B = 4.0 T
The torque acting on the coil is calculated as, τ = NIAB sinθ
τ = 100 × 80 × 10-3 × π × (0.50)2 × 4.0 × sin 30º
τ = 4π N m
Hence, the magnitude of torque acting on the coil is 4π N m.
Last updated on May 26, 2025
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