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: Understanding the Concept:
When a loop rotates in a magnetic field, the magnetic flux through it changes continuously. According to Faraday's law, this change induces an alternating electromotive force (EMF).

Step 2: Key Formula or Approach:
Maximum induced emf \(\varepsilon_0 = NBA\omega\). Here \(N=1\).

Step 3: Detailed Explanation:
1. Area \(A = (0.1 \text{ m})^2 = 0.01 \text{ m}^2\). 2. Magnetic field \(B = 0.2 \text{ T}\). 3. Angular velocity \(\omega\): \[ f = 60 \text{ rpm} = \frac{60}{60} \text{ rps} = 1 \text{ Hz} \] \[ \omega = 2\pi f = 2\pi(1) = 2\pi \text{ rad/s} \] 4. Induced EMF (instantaneous) is \(\varepsilon = BA\omega \sin(\omega t)\). The peak value is: \[ \varepsilon_0 = 0.2 \times 0.01 \times 2\pi = 0.002 \times 2\pi = 0.004\pi \text{ V} \] \[ \varepsilon_0 \approx 0.01256 \text{ V} \]

Step 4: Final Answer:
The maximum induced emf is \(0.004\pi\) V (or approximately \(1.26 \times 10^{-2}\) V).