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:
Solution
<p>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:</p>
<p>$ \text{emf} = B \cdot L \cdot v $</p>
<p>where:</p>
<ul>
<li>(B) is the magnetic field strength,</li>
<li>(L) is the length of the rod, and</li>
<li>(v) is the velocity of the rod.</li>
</ul>
<p>In this case, we are given the emf, (B), and (L), and we need to solve for (v). Rearranging the equation gives:</p>
<p>$ v = \frac{\text{emf}}{B \cdot L} $</p>
<p>Substituting the given values:</p>
<p>$ v = \frac{0.08 \, \text{V}}{0.4 \, \text{T} \times0.1 \, \text{m}} = 2 \, \text{m/s} $</p>
About this question
Subject: Physics · Chapter: Electromagnetic Induction · Topic: Faraday's Laws
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