Question

A metal loop of area \[ 10\,\text{cm}^2 \] is placed in a region such that its area vector points along \(\hat{k}\). The region contains a uniform magnetic field of magnitude \(1.73\,\text{T}\) that points in the direction \[ \hat{i}+\hat{j}+\hat{k}. \] When the magnetic field is switched off, the field decreases to zero at a steady rate in \(10\,\text{s}\), then the magnitude of emf induced in the loop is:

• A. \(0.10\,\text{mV}\)
• B. \(0.17\,\text{mV}\)
• C. \(1\,\text{mV}\)
• D. \(1.7\,\text{mV}\)

Hint:

Magnetic flux depends only on the component of magnetic field normal to the loop: \[ \Phi=\vec{B}\cdot \vec{A}. \] Always resolve the magnetic field along the area vector direction.

Answer 1

The Correct Option is A

Step 1: Determine the component of magnetic field normal to the loop.
The magnetic field direction is \[ \hat{i}+\hat{j}+\hat{k} \] Magnitude of this vector: \[ \sqrt{1^2+1^2+1^2}=\sqrt{3} \] Therefore, the unit vector along magnetic field is \[ \frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}} \] Since the total magnetic field magnitude is \[ 1.73\,\text{T}, \] the component along \[ \hat{k} \] is \[ B_k=1.73\left(\frac{1}{\sqrt{3}}\right) \] \[ B_k\approx \frac{1.73}{1.732} \] \[ B_k\approx 1\,\text{T} \]

Step 2: Convert area into SI unit.
Given, \[ A=10\,\text{cm}^2 \] \[ A=10\times10^{-4}\,\text{m}^2 \] \[ A=10^{-3}\,\text{m}^2 \]

Step 3: Find the initial magnetic flux.
Magnetic flux is \[ \Phi=BA \] \[ \Phi=1\times10^{-3} \] \[ \Phi=10^{-3}\,\text{Wb} \]

Step 4: Use Faraday’s law.
The magnetic field becomes zero uniformly in \[ 10\,\text{s} \] Hence, \[ \mathcal{E}=\frac{\Delta\Phi}{\Delta t} \] \[ \mathcal{E}=\frac{10^{-3}}{10} \] \[ \mathcal{E}=10^{-4}\,\text{V} \] \[ \mathcal{E}=0.10\,\text{mV} \]

Step 5: Final conclusion.
Hence, the induced emf is \[ \boxed{0.10\,\text{mV}} \]