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Difficulty: MediumMagnetic Force and Electromagnetism

A rectangular coil consisting of 5050 turns of wire has dimensions of 0.10 m0.10\text{ m} by 0.05 m0.05\text{ m} and carries a steady electric current of 2.0 A2.0\text{ A}. The coil is suspended in a uniform magnetic field of flux density 0.40 T0.40\text{ T}. If the normal to the plane of the coil makes an angle of 30∘30^\circ with the direction of the magnetic field, what is the magnitude of the torque exerted on the coil?

  1. 0.10 N⋅m0.10\text{ N}\cdot\text{m}Answer
  2. B
    0.17 N⋅m0.17\text{ N}\cdot\text{m}
  3. C
    0.20 N⋅m0.20\text{ N}\cdot\text{m}
  4. D
    3.00 N⋅m3.00\text{ N}\cdot\text{m}

Answer

The magnitude of the torque exerted on the coil is 0.10 N⋅m0.10\text{ N}\cdot\text{m}.
The magnitude of torque on a current-carrying coil situated in a uniform magnetic field is given by τ=NIABsin⁡θ\tau = N I A B \sin\theta, where θ\theta is the angle between the normal vector of the coil and the magnetic field vector. Substituting N=50N = 50, I=2.0 AI = 2.0\text{ A}, A=0.005 m2A = 0.005\text{ m}^2, B=0.40 TB = 0.40\text{ T}, and θ=30∘\theta = 30^\circ yields τ=50×2.0×0.005×0.40×0.5=0.10 N⋅m\tau = 50 \times 2.0 \times 0.005 \times 0.40 \times 0.5 = 0.10\text{ N}\cdot\text{m}.

Step-by-Step Solution

1
Calculate the area AA of the rectangular coil.
A=length×width=0.10 m×0.05 m=0.005 m2A = \text{length} \times \text{width} = 0.10\text{ m} \times 0.05\text{ m} = 0.005\text{ m}^2
The torque on a coil depends on its total surface area.
2
Identify the given values and formula for magnetic torque.
Formula: τ=NIABsin⁡θ\tau = N I A B \sin\theta, where N=50N = 50, I=2.0 AI = 2.0\text{ A}, A=0.005 m2A = 0.005\text{ m}^2, B=0.40 TB = 0.40\text{ T}, and θ=30∘\theta = 30^\circ.
When θ\theta is measured relative to the normal of the coil's plane, the sine component determines the effective perpendicular force arm.
3
Substitute the values into the formula and solve for torque τ\tau.
τ=50×2.0 A×0.005 m2×0.40 T×sin⁡30∘=100×0.002×0.5=0.10 N⋅m\tau = 50 \times 2.0\text{ A} \times 0.005\text{ m}^2 \times 0.40\text{ T} \times \sin 30^\circ = 100 \times 0.002 \times 0.5 = 0.10\text{ N}\cdot\text{m}
Direct evaluation yields the final magnetic torque.

Key Concept

Torque on a Current-Carrying Loop in a Uniform Magnetic Field
Estimated Time:1m 30s
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Topics

Magnetic Force and Electromagnetism

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