Question

A square loop of side \( 10 \, \text{cm} \) and resistance \( 0.7 \, \Omega \) is placed vertically in the east-west plane. A uniform magnetic field of \( 0.20 \, \text{T} \) is set up across the plane in the northeast direction. The magnetic field is decreased to zero in \( 1 \, \text{s} \) at a steady rate. Then, the magnitude of the induced emf is \( \sqrt{x} \times 10^{-3} \, \text{V} \). The value of \( x \) is _____.

Answer 1

Correct Answer: 2

Step 1: Calculate Area Vector of the Square Loop:

- Side of square = 10 cm = 0.1 m

- Area \( A = (0.1)^2 = 0.01 \, \text{m}^2 \). Since the loop is placed in the east-west plane, the area vector \( \vec{A} \) is along the \( \hat{j} \) direction:

\[ \vec{A} = 0.01 \, \hat{j} \, \text{m}^2 \]

Step 2: Calculate the Magnetic Field Vector \( \vec{B} \):

- The magnetic field \( B = 0.20 \, \text{T} \) is directed at a \( 45^\circ \) angle in the northeast direction, so:

\[ \vec{B} = \frac{0.20}{\sqrt{2}} \, \hat{i} + \frac{0.20}{\sqrt{2}} \, \hat{j} \]

- Simplify:

\[ \vec{B} = 0.1414 \, \hat{i} + 0.1414 \, \hat{j} \, \text{T} \]

Step 3: Calculate the Magnetic Flux \( \Phi \):

\[ \Phi = \vec{B} \cdot \vec{A} = (0.1414 \, \hat{i} + 0.1414 \, \hat{j}) \cdot (0 \, \hat{i} + 0.01 \, \hat{j}) \]

\[ \Phi = 0.1414 \times 0.01 = 0.001414 \, \text{Wb} \]

Step 4: Calculate Induced EMF (\( \varepsilon \)):

- The magnetic field is reduced to zero in \( \Delta t = 1 \, \text{s} \), so:

\[ \varepsilon = -\frac{\Delta \Phi}{\Delta t} = -\frac{0.001414 - 0}{1} = 0.001414 \, \text{V} = \sqrt{2} \times 10^{-3} \, \text{V} \]

Step 5: Determine \( x \):

- Since \( \varepsilon = \sqrt{x} \times 10^{-3} \, \text{V} \), we have \( x = 2 \).

So, the correct answer is: \(x = 2\)