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

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: $x = 2$