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Step-by-Step Solution
1. Understand the Problem
A circular loop is placed in a uniform magnetic field of magnitude $0.025\,\text{T}$, with its plane perpendicular to the field. The loop’s radius $r$ decreases at a constant rate of $1 \times 10^{-3}\,\text{m s}^{-1}$. We need to find the induced emf when $r = 2\,\text{cm}$ (i.e., $2 \times 10^{-2}\,\text{m}$).
2. List the Known Quantities
Magnetic field, $B = 0.025\,\text{T}$
Rate of change of radius, $\frac{dr}{dt} = 1 \times 10^{-3}\,\text{m s}^{-1}$
Radius at the instant considered, $r = 2 \times 10^{-2}\,\text{m}$
3. Write the Expression for Magnetic Flux
The magnetic flux $ \phi $ through the loop (with its plane perpendicular to the field) is given by:
$ \phi = B \times \text{Area} = B \times \pi r^{2}. $
4. Determine the Induced emf
The induced emf $e$ is given by the magnitude of the rate of change of magnetic flux:
$ |e| = \left|\frac{d \phi}{dt}\right| = \left|\frac{d}{dt}\bigl(B \pi r^{2}\bigr)\right|. $
Since $B$ and $\pi$ are constants, we have:
$ |e| = B \pi \left|\frac{d}{dt}(r^{2})\right| = B \pi \left|2r \frac{dr}{dt}\right| = 2\pi B r \frac{dr}{dt}. $
5. Substitute the Numerical Values
$ B = 0.025\,\text{T}, \quad r = 2 \times 10^{-2}\,\text{m}, \quad \frac{dr}{dt} = 1 \times 10^{-3}\,\text{m s}^{-1}. $
$ |e| = 2 \times \pi \times (0.025) \times (2 \times 10^{-2}) \times (1 \times 10^{-3}). $
First, multiply the constants without $\pi$:
$ 2 \times 0.025 = 0.05, \quad 0.05 \times (2 \times 10^{-2}) = 0.05 \times 0.02 = 0.001, \quad 0.001 \times 10^{-3} = 10^{-6}. $
Incorporating $\pi$ gives:
$ |e| = \pi \times 10^{-6}\,\text{V}. $
6. Final Answer
$ |e| = \pi \mu \text{V} $ (where $\mu = 10^{-6}$).
