Question

A uniform magnetic field \(\vec{B}\) is perpendicular to the plane of a circular loop of diameter \(10\ \text{cm}\) formed from wire of diameter \(2\ \text{mm}\) and resistivity \(2\times 10^{-8}\ \Omega\text{m}\). If a current of \(11\ \text{A}\) is to be induced in the loop, then the rate at which \(\vec{B}\) is to be changed is

• A. \(2.8\ \text{Ts}^{-1}\)
• B. \(1.4\ \text{Ts}^{-1}\)
• C. \(3.2\ \text{Ts}^{-1}\)
• D. \(2.4\ \text{Ts}^{-1}\)

Hint:

For a conducting loop in a changing magnetic field: \[ \varepsilon=A\frac{dB}{dt} \] and \[ \varepsilon=IR \] First calculate resistance of the wire, then induced emf, and finally \(\frac{dB}{dt}\).

Answer 1

The Correct Option is A

Step 1: Find the resistance of the circular wire.
Resistance of wire is \[ R=\rho \frac{l}{A} \] Diameter of circular loop: \[ D=10\ \text{cm}=0.10\ \text{m} \] Length of wire: \[ l=\pi D=\pi(0.10) \] \[ l=0.10\pi\ \text{m} \] Diameter of wire: \[ d=2\ \text{mm}=2\times 10^{-3}\ \text{m} \] Radius of wire: \[ r=1\times 10^{-3}\ \text{m} \] Area of cross-section: \[ A=\pi r^2 \] \[ A=\pi(10^{-3})^2 \] \[ A=\pi\times 10^{-6}\ \text{m}^2 \] Now, \[ R=2\times 10^{-8}\times \frac{0.10\pi}{\pi\times 10^{-6}} \] \[ R=2\times 10^{-8}\times 10^5 \] \[ R=2\times 10^{-3}\ \Omega \]

Step 2: Find the induced emf.
Using Ohm's law, \[ \varepsilon=IR \] Given: \[ I=11\ \text{A} \] So, \[ \varepsilon=11\times 2\times 10^{-3} \] \[ \varepsilon=22\times 10^{-3} \] \[ \varepsilon=0.022\ \text{V} \]

Step 3: Use Faraday's law of electromagnetic induction.
Since magnetic field is perpendicular to the plane of loop, \[ \varepsilon=A_{\text{loop}}\frac{dB}{dt} \] Radius of loop: \[ R_{\text{loop}}=\frac{10}{2}\ \text{cm}=5\ \text{cm}=0.05\ \text{m} \] Area of loop: \[ A_{\text{loop}}=\pi R_{\text{loop}}^2 \] \[ A_{\text{loop}}=\pi(0.05)^2 \] \[ A_{\text{loop}}=0.0025\pi\ \text{m}^2 \]

Step 4: Find the rate of change of magnetic field.
\[ \frac{dB}{dt}=\frac{\varepsilon}{A_{\text{loop}}} \] \[ \frac{dB}{dt}=\frac{0.022}{0.0025\pi} \] Using \[ \pi\approx 3.14, \] \[ \frac{dB}{dt}=\frac{0.022}{0.00785} \] \[ \frac{dB}{dt}\approx 2.8\ \text{Ts}^{-1} \]

Step 5: Final conclusion.
Hence, the required rate of change of magnetic field is \[ \boxed{2.8\ \text{Ts}^{-1}} \]