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 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 ______
Updated On: Mar 19, 2025
- $20 V$
- $16 V$
- $8 V$
- $12 V$
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The Correct Option is B
Approach Solution - 1
The correct answer is (B) : $16 V$
Induced emf across the ends
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Approach Solution -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:
\[
\varepsilon = 2 \, \text{T} \times 8 \, \text{m/s} \times 1 \, \text{m} = 16 \, \text{V}
\]
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Concepts Used:
Electromagnetic Induction
Electromagnetic Induction is a current produced by the voltage production due to a changing magnetic field. This happens in one of the two conditions:-
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- When the conductor constantly moves in a stationary field.
Formula:
The electromagnetic induction is mathematically represented as:-
e=N × d∅.dt
Where
- e = induced voltage
- N = number of turns in the coil
- Φ = Magnetic flux (This is the amount of magnetic field present on the surface)
- t = time
Applications of Electromagnetic Induction
- Electromagnetic induction in AC generator
- Electrical Transformers
- Magnetic Flow Meter