A wire of length \( L \) and resistance \( R \) is falling vertically through Earth's horizontal magnetic field \( B \). What is the current induced in the wire when it has fallen a height \( L \)? (Take acceleration due to gravity as \( g \))
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When a wire falls through a magnetic field, the induced current depends on the velocity of the wire, which is influenced by the gravitational potential energy being converted to kinetic energy.
The current induced in the wire is due to the motion of the wire through the magnetic field. According to Faraday's law, the induced emf (\( \epsilon \)) is given by:
\[
\epsilon = BvL
\]
Where \( v \) is the velocity of the wire as it falls under the influence of gravity. Using the equation \( v^2 = 2gL \), the velocity is:
\[
v = \sqrt{2gL}
\]
Thus, the induced emf becomes:
\[
\epsilon = B\sqrt{2gL}L
\]
The current \( I \) is given by Ohm's law:
\[
I = \frac{\epsilon}{R} = \frac{B\sqrt{2gL}L}{R}
\]
Thus, the induced current is:
\[
I = \frac{BL\sqrt{2gL}}{R}
\]
