Question:
A rectangular coil of $100$ turns and size $0.1 \times 0.05\, m$ is placed perpendicular to a magnetic field of $0.1\, T$. The induced emf when the field drops to $0.05\, T$ in $0.05\, s$ is
A rectangular coil of $100$ turns and size $0.1 \times 0.05\, m$ is placed perpendicular to a magnetic field of $0.1\, T$. The induced emf when the field drops to $0.05\, T$ in $0.05\, s$ is
Updated On: Jun 20, 2022
- 0.5 V
- 1.0 V
- 1.5 V
- 2.0 V
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The Correct Option is A
Solution and Explanation
The induced emf
$e=-N \frac{\Delta \phi}{\Delta t}$
or $e=-\frac{N\left(B_{2}-B_{1}\right) A}{\Delta t}$
or $e =-\frac{100(0.1-0.05) \times 0.1 \times 0.05}{0.05}$
$e =0.5\, V$
$e=-N \frac{\Delta \phi}{\Delta t}$
or $e=-\frac{N\left(B_{2}-B_{1}\right) A}{\Delta t}$
or $e =-\frac{100(0.1-0.05) \times 0.1 \times 0.05}{0.05}$
$e =0.5\, V$
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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:-
- When we place the conductor in a changing magnetic field.
- 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