Question

A wire of length $1 m$ moving with velocity $8 m / s$ at right angles to a magnetic field of $2 T$ The magnitude of induced emf, between the ends of wire will be ______

• A. $20 V$
• B. $16 V$
• C. $8 V$
• D. $12 V$

Answer 1

The Correct Option is B

The correct answer is (B) : $16 V$

Induced emf across the ends $=Bv\ell$
$=2\times 8\times 1=16V$

Answer 2

The induced electromotive force (emf), denoted by \( \varepsilon \), in a wire moving perpendicularly through a magnetic field is given by the formula: \[ \varepsilon = Bvl \] where \( B \) represents the magnetic field strength, \( v \) is the velocity of the wire, and \( l \) is the length of the wire. Substituting the given values: \[ \varepsilon = 2 \, \text{T} \times 8 \, \text{m/s} \times 1 \, \text{m} = 16 \, \text{V} \]