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Question 15

A wire of length 1 m moving with velocity 8 m s$$^{-1}$$ at right angles to a magnetic field of 2 T. The magnitude of induced emf, between the ends of wire will be _____.

Solution

The induced emf in a wire moving perpendicular to a magnetic field is given by:

$$\varepsilon=Blv$$

Where:

B=2 T
l=1 m
v=8 ms−1

Substituting:

$$\varepsilon=2\times1\times8$$

$$\varepsilon=16\text{ V}$$

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