Electric Motor Physics Example 2

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Example 2

medium
A motor coil has 200200 turns, area 0.01 m20.01 \text{ m}^2, carries 3 A3 \text{ A}, and is in a 0.5 T0.5 \text{ T} field. What is the maximum torque on the coil?

Solution

  1. 1
    Torque on a current loop: τ=NIABsin⁡θ\tau = NIAB\sin\theta.
  2. 2
    Maximum torque occurs when θ=90°\theta = 90° (sin⁡90°=1\sin 90° = 1).
  3. 3
    τmax⁡=NIAB=200×3×0.01×0.5=3 N m\tau_{\max} = NIAB = 200 \times 3 \times 0.01 \times 0.5 = 3 \text{ N m}

Answer

τmax⁡=3 N m\tau_{\max} = 3 \text{ N m}
A motor works because a current-carrying coil in a magnetic field experiences a torque. The commutator reverses the current direction each half-turn to keep the coil spinning continuously.

About Electric Motor

A device that converts electrical energy into mechanical energy (rotational motion) by exploiting the force exerted on a current-carrying conductor placed in a magnetic field.

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