Question

A rectangular loop of area vector \(\vec{A} = [\hat{i} + \sqrt{3}\,\hat{j}] \times 10^{-2}\,\text{m}^2\) is placed in a magnetic field \(\vec{B} = [4.0t^2 + 2.0]\,\hat{i}\), where \(B\) is in tesla and \(t\) in seconds. At \(t = 1.0\) s, the magnitude of induced emf in the loop is:

• A. 80 mV
• B. 0.11 V
• C. 0.22 V
• D. 0.8 V

Hint:

Flux is \(\vec{B}\cdot\vec{A}\), and only the x-component of the area counts. Differentiate the flux and put \(t = 1\) s.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
By Faraday law, \(\varepsilon = -\dfrac{d\Phi}{dt}\), where \(\Phi = \vec{B} \cdot \vec{A}\) is the magnetic flux through the loop.
Only the component of the area vector along the field counts.

Step 2: Find the flux:
\(\vec{A} = (\hat{i} + \sqrt{3}\,\hat{j}) \times 10^{-2}\ \text{m}^2\) and \(\vec{B} = (4.0t^2 + 2.0)\,\hat{i}\ \text{T}\).
\[ \Phi = \vec{B} \cdot \vec{A} = (4.0t^2 + 2.0)(1 \times 10^{-2}) \ \text{Wb} \]
The \(\hat{j}\) part of \(\vec{A}\) gives zero in the dot product with \(\hat{i}\).

Step 3: Differentiate:
\[ \frac{d\Phi}{dt} = 8.0\,t \times 10^{-2}\ \text{V} \]
At \(t = 1.0\) s: \(|\varepsilon| = 8.0 \times 10^{-2} = 0.08\ \text{V} = 80\ \text{mV}\).

Step 4: Why the other options are wrong:
(C) 0.22 V: this comes from wrongly adding both components, \(8(1 + \sqrt{3}) \times 10^{-2} \approx 0.22\) V. The \(\hat{j}\) component must not be included.
(D) 0.8 V is ten times too large. (B) 0.11 V does not follow from any correct use of the data.

Final Answer:
The magnitude of the induced emf at \(t = 1.0\) s is 80 mV. This is option (A). \[ \boxed{\varepsilon = 80\ \text{mV}} \]