Question 11: Electromagnetic Induction

Question

A conducting circular loop is placed in a uniform magnetic field $$\mathrm{B}=0.125 \mathrm{~T}$$ with its plane perpendicular to the loop. If the radius of the loop is made to shrink at a constant rate of $$2 \mathrm{~mm} \mathrm{~s}^{-1}$$, then the induced emf when the radius is $$4 \mathrm{~cm}$$ is

Accepted Answer

First, we need to find the rate at which the area of the loop is changing since the induced emf in the loop is related to the change of magnetic flux through the loop.
The magnetic flux $$\Phi$$ through the loop is given by:
$$\Phi = B \cdot A$$
where
$$B$$ is the magnetic field
and
$$A$$ is the area of the loop.
The area $$A$$ of a circular loop with radius $$r$$ is given by:
$$A = \pi r^2$$
The rate of change of the area $$A$$ with respect to time $$t$$ is:
$$\frac{dA}{dt} = \pi \frac{d}{dt} (r^2) = \pi \cdot 2r \cdot \frac{dr}{dt}$$
We are given:
$$\frac{dr}{dt} = -2 \mathrm{~mm} \mathrm{~s}^{-1} = -2 	imes 10^{-3} \mathrm{~m} \mathrm{~s}^{-1}$$
$$r = 4 \mathrm{~cm} = 0.04 \mathrm{~m}$$
Substituting these values in, we get:
$$\frac{dA}{dt} = \pi \cdot 2r \cdot \frac{dr}{dt}$$
$$\frac{dA}{dt} = \pi \cdot 2 \cdot 0.04 \cdot -2 	imes 10^{-3}$$
$$\frac{dA}{dt} = \pi \cdot 2 \cdot 0.04 \cdot -2 \cdot 10^{-3}$$
$$\frac{dA}{dt} = -1.6 	imes 10^{-4} \pi \mathrm{~m}^2 \mathrm{~s}^{-1}$$
The induced emf $$\varepsilon$$ is given by Faraday's law of induction:
$$\varepsilon = - \frac{d\Phi}{dt}$$
Since$$ \Phi = B \cdot A$$, we have:
$$\varepsilon = - B \cdot \frac{dA}{dt}$$
Substituting the values of $$B$$ and $$\frac{dA}{dt}$$, we get:
$$\varepsilon = -0.125 \mathrm{~T} \cdot  -1.6 	imes 10^{-4} \pi \mathrm{~m}^2 \mathrm{~s}^{-1}$$
$$\varepsilon = 0.125 \cdot 1.6 	imes 10^{-4} \pi$$
$$\varepsilon = 2 	imes 10^{-5} \pi \mathrm{~V}$$
Since the question asks for the answer in $$\mu V$$ (microvolts), we convert the emf:
$$\varepsilon = 20 \pi \mu \mathrm{V}$$
Thus, the correct answer is:
Option B: $$20 \pi \mu V$$

Answer

$$0.52 \pi \mu V$$

Answer

$$20 \pi \mu V$$

Answer

$$\frac{2}{3}\mu V$$

Answer

$$\frac{3 \pi}{2} \mu V$$

Electromagnetic Induction

A conducting circular loop is placed in a uniform magnetic field $\mathrm{B}=0.125 \mathrm{~T} with its plane perpendicular to the loop. If the radius of the loop is made to shrink at a constant rate of 2 \mathrm{~mm} \mathrm{~s}^{-1}, then the induced emf when the radius is 4 \mathrm{~cm}$ is

Choose the correct answer: