JEE MAIN - Physics (2023 - 8th April Evening Shift - No. 2)
An emf of $$0.08 \mathrm{~V}$$ is induced in a metal rod of length $$10 \mathrm{~cm}$$ held normal to a uniform magnetic field of $$0.4 \mathrm{~T}$$, when moves with a velocity of:
$$20 \mathrm{~ms}^{-1}$$
$$2 \mathrm{~ms}^{-1}$$
$$3.2 \mathrm{~ms}^{-1}$$
$$0.5 \mathrm{~ms}^{-1}$$
Explanation
The emf induced in a rod moving through a magnetic field is given by Faraday's law of electromagnetic induction, specifically, in the form of motional emf, which states that:
$ \text{emf} = B \cdot L \cdot v $
where:
- (B) is the magnetic field strength,
- (L) is the length of the rod, and
- (v) is the velocity of the rod.
In this case, we are given the emf, (B), and (L), and we need to solve for (v). Rearranging the equation gives:
$ v = \frac{\text{emf}}{B \cdot L} $
Substituting the given values:
$ v = \frac{0.08 \, \text{V}}{0.4 \, \text{T} \times0.1 \, \text{m}} = 2 \, \text{m/s} $
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