Question

A magnetic field induction is changing in magnitude at a constant rate \(\frac{dB}{dt}\). The plane of a circular loop of radius \(r\) of copper wire (mass \(m\) and radius \(a\)) is placed perpendicular to the field. The induced current in the loop is {\(\rho\) = specific resistance, \(\delta\) = density of copper}

• A. \(\frac{2\pi \rho \delta }{m}\frac{dB}{dt}\)
• B. \(\frac{m}{2\pi \rho \delta }\frac{dB}{dt}\)
• C. \(\frac{m}{4\pi \rho \delta }\frac{dB}{dt}\)
• D. \(\frac{4\pi \rho \delta }{m}\frac{dB}{dt}\)

Hint:

Find the emf, the resistance of the wire, and express the wire length through its mass.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Induced emf: \(\varepsilon=\pi r^2\dfrac{dB}{dt}\).

Step 2: Resistance
The wire has length \(2\pi r\) and cross-section \(\pi a^2\), so \(R=\dfrac{\rho\,2\pi r}{\pi a^2}=\dfrac{2\rho r}{a^2}\).

Step 3: Use the mass
\(m=\delta\,(2\pi r)(\pi a^2)\), so \(ra^2=\dfrac{m}{2\pi^2\delta}\).

Step 4: Calculation
\(I=\dfrac{\varepsilon}{R}=\dfrac{\pi r^2a^2}{2\rho r}\dfrac{dB}{dt}=\dfrac{\pi\,r a^2}{2\rho}\dfrac{dB}{dt}=\dfrac{\pi}{2\rho}\cdot\dfrac{m}{2\pi^2\delta}\dfrac{dB}{dt}\).

Step 5: Result
\(I=\dfrac{m}{4\pi\rho\delta}\dfrac{dB}{dt}\).

Final Answer:
The induced current is m divided by 4 pi rho delta times dB/dt. \[ \boxed{\text{(C) }\dfrac{m}{4\pi \rho \delta }\dfrac{dB}{dt}} \]