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: \(x = 2\)

Answer 2

Step 1: Given data.

\[ \text{Side of square} = 10\,\text{cm} = 0.1\,\text{m} \] \[ \text{Magnetic field, } B = 0.20\,\text{T} \] \[ \text{Time interval, } \Delta t = 1\,\text{s} \] \[ \text{Resistance } R = 0.7\,\Omega \]

Step 2: Find component of \( B \) perpendicular to the loop.

The loop is in the east–west vertical plane, and the magnetic field is along the northeast direction. Thus, the magnetic field makes an angle of \( 45^\circ \) with the normal to the loop.

So, the component of the magnetic field perpendicular to the loop is: \[ B_\perp = B \cos 45^\circ = 0.20 \times \frac{1}{\sqrt{2}} = 0.1414\,\text{T} \]

Step 3: Magnetic flux through the loop.

\[ \phi = B_\perp \times A \] \[ A = (\text{side})^2 = (0.1)^2 = 0.01\,\text{m}^2 \] \[ \phi = 0.1414 \times 0.01 = 1.414 \times 10^{-3}\,\text{Wb} \]

Step 4: Rate of change of magnetic flux = induced emf.

\[ \varepsilon = \frac{\Delta \phi}{\Delta t} = \frac{1.414 \times 10^{-3}}{1} = 1.414 \times 10^{-3}\,\text{V} \] \[ \varepsilon \approx 1.4 \times 10^{-3}\,\text{V} \]

Step 5: Round to the given format.

\[ \varepsilon = x \times 10^{-3}\,\text{V} \Rightarrow x = 1.4 \approx 2 \]

Final Answer:

\[ \boxed{x = 2} \]