Question

A metallic circular loop is placed with its plane perpendicular to a uniform magnetic field of $3\text{ T}$. If the radius of the loop decreases at a constant rate of $2\text{ mm s}^{-1}$, what will be the induced emf in the loop when the radius of the loop becomes $5\text{ cm}$?

• A. $1.5 \times 10^{-6}\text{ V}$
• B. $1.89 \times 10^{-4}\text{ V}$
• C. $1.84 \times 10^{-6}\text{ V}$
• D. $0.75 \times 10^{-6}\text{ V}$

Hint:

Always keep your variables in strict SI units! Convert radius from $\text{cm}$ to $\text{m}$ and rates from $\text{mm/s}$ to $\text{m/s}$ right away to guarantee your final answer comes out correctly in Volts ($\text{V}$).

Answer 1

The Correct Option is B

Concept: According to Faraday's Law of Induction, a change in magnetic flux ($\phi$) linked with a conducting loop induces an electromotive force (emf) given by: \[ e = -\frac{d\phi}{dt} \] The magnetic flux through a flat loop perpendicular to a uniform field $B$ is $\phi = BA = B(\pi r^2)$. If the radius changes over time, the rate of change of area induces an electrical response.

Step 1: Differentiate the flux expression with respect to time.
Given that $\phi = B \cdot \pi r^2$, differentiating both sides gives: \[ e = \left| \frac{d\phi}{dt} \right| = B \cdot \pi \cdot \frac{d}{dt}(r^2) = B \cdot \pi \cdot 2r \frac{dr}{dt} \]

Step 2: Substitute the given parameters into the derived equation.
We are given:
• Magnetic field, $B = 0.3\text{ T}$
• Instantaneous radius, $r = 5\text{ cm} = 0.05\text{ m}$
• Rate of decrease of radius, $\frac{dr}{dt} = 2\text{ mm s}^{-1} = 2 \times 10^{-3}\text{ m s}^{-1}$ Plugging these into the magnitude equation: \[ e = 0.3 \times \pi \times 2(0.05) \times (2 \times 10^{-3}) \] \[ e = 0.3 \times \pi \times 0.1 \times (2 \times 10^{-3}) = 0.06 \times \pi \times 10^{-3} = 6 \times 10^{-5} \times \pi \] Taking $\pi \approx 3.1416$: \[ e = 6 \times 3.1416 \times 10^{-5} \approx 18.85 \times 10^{-5}\text{ V} = 1.885 \times 10^{-4}\text{ V} \approx 1.89 \times 10^{-4}\text{ V} \]