Question

A square loop of side 10 cm, free to rotate about a vertical axis coinciding with its one arm, is initially held perpendicular to a uniform horizontal magnetic field of 0.2 T. If it is rotated at the uniform speed of 60 rpm, find the emf induced in the loop.

Hint:

60 rpm is exactly 1 revolution per second. In many AC generator problems, this corresponds to a frequency of 50Hz or 60Hz; here it is simply 1Hz.

Answer 1

Step 1: Principle involved.
When a conducting loop rotates in a uniform magnetic field, the magnetic flux linked with it changes with time. According to Faraday’s law, a changing flux induces an alternating emf.
Step 2: Formula for maximum induced emf.
\[ \varepsilon_0 = NBA\omega \] Here, \(N = 1\).
Step 3: Calculate required quantities.
Area of the square loop: \[ A = (0.1)^2 = 0.01 \, \text{m}^2 \] Magnetic field: \[ B = 0.2 \, \text{T} \] Convert rotational speed into angular velocity: \[ f = 60 \, \text{rpm} = \frac{60}{60} = 1 \, \text{Hz} \] \[ \omega = 2\pi f = 2\pi \times 1 = 2\pi \, \text{rad/s} \]
Step 4: Determine maximum emf.
\[ \varepsilon_0 = 0.2 \times 0.01 \times 2\pi \] \[ \varepsilon_0 = 0.002 \times 2\pi = 0.004\pi \, \text{V} \] \[ \varepsilon_0 \approx 0.01256 \, \text{V} \]
Final Answer:
Maximum induced emf \( = 0.004\pi \, \text{V} \)
or approximately \( 1.26 \times 10^{-2} \, \text{V} \)