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Q. A wire of length $50 \,cm$ moves with a velocity of $300\, m - \min ^{-1}$, perpendicular to a magnetic field. If the emf induced in the wire is $2\, V$, the magnitude of the field (in tesla) is

Electromagnetic Induction

Solution:

Emf induced in the wire is given by, $\varepsilon=B l v$
Given, $l=50\, cm =0.5 \, m $
$v=300 \, m -\min ^{-1}=\frac{300}{60}=5 \, ms ^{-1}$
$e=2\, V$
Magnetic field, $B=\frac{\varepsilon}{l v}=\frac{2}{0.5 \times 5}=0.80\, T$