Question

A square loop of side \(2\,\text{cm}\) is placed in a time varying magnetic field with magnitude \(B=0.4\sin(300t)\) Tesla. The normal to the plane of loop makes an angle \(60^\circ\) with the field. The maximum induced emf produced in the loop is _____ mV.

• A. \(12\)
• B. \(18\)
• C. \(21\)
• D. \(24\)

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the peak value of the induced electromotive force (emf) in a loop when subjected to a changing magnetic field.
Step 2: Key Formula or Approach:
According to Faraday's law of induction, the induced emf \(e\) is:
\[ e = -\frac{d\Phi}{dt} \]
Magnetic flux \(\Phi = B \cdot A \cdot \cos\theta\).
Step 3: Detailed Explanation:
Area of square loop \(A = (0.02 \text{ m})^2 = 4 \times 10^{-4} \text{ m}^2\).
Angle \(\theta = 60^\circ \Rightarrow \cos(60^\circ) = 0.5\).
Flux \(\Phi = [0.4 \sin(300t)] \times [4 \times 10^{-4}] \times 0.5\)
\[ \Phi = 0.8 \times 10^{-4} \sin(300t) \]
Differentiating with respect to time to find emf:
\[ e = \left| \frac{d\Phi}{dt} \right| = 0.8 \times 10^{-4} \times 300 \cos(300t) \]
\[ e = 0.024 \cos(300t) \text{ Volts} \]
The maximum value occurs when \(\cos(300t) = 1\):
\[ e_{max} = 0.024 \text{ V} = 24 \text{ mV} \]
Step 4: Final Answer:
The maximum induced emf is \(24 \text{ mV}\).