Question:

A wire is cut across a flux of \( 0.2 \times 10^{-2} \) weber in 0.12 seconds. What is the e.m.f. induced in the wire?

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The induced emf is proportional to the rate of change of magnetic flux through the wire, as given by Faraday’s law of induction.
Updated On: Feb 3, 2026
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Solution and Explanation

Step 1: Formula for induced emf.
The induced emf \( \varepsilon \) is given by Faraday's law of electromagnetic induction: \[ \varepsilon = -\frac{d\Phi_B}{dt} \] where \( \Phi_B \) is the magnetic flux and \( \frac{d\Phi_B}{dt} \) is the rate of change of flux. Step 2: Substituting known values.
Given that the flux \( \Phi_B = 0.2 \times 10^{-2} \, \text{weber} \) and the time interval \( \Delta t = 0.12 \, \text{seconds} \), the rate of change of flux is: \[ \frac{d\Phi_B}{dt} = \frac{0.2 \times 10^{-2}}{0.12} = 1.67 \times 10^{-3} \, \text{weber/second} \] Step 3: Calculating the induced emf.
The induced emf is: \[ \varepsilon = -1.67 \times 10^{-3} \, \text{V} = -1.67 \, \text{mV} \] The negative sign indicates the direction of the induced emf according to Lenz’s law, but the magnitude of the emf is \( 1.67 \, \text{mV} \). Step 4: Conclusion.
Thus, the induced emf in the wire is \( 1.67 \, \text{mV} \).
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