A small piece of metal wire is dragged across the gap between the poles of a magnet is 0.4 s. If the change in magnetic flux in the wire is 8× 10-4 Wb, then emf induced in the wire is:

Question

A small piece of metal wire is dragged across the gap between the poles of a magnet is 0.4 s. If the change in magnetic flux in the wire is 8× 10-4 Wb, then emf induced in the wire is: (A)  8× 10-3 V  (B)  6× 10-3 V  (C)  4× 10-3 V  (D)  2× 10-3 V

Tardigrade Admin · Accepted Answer

From Faraday's law of electromagnetic induction, when the magnetic flux through a circuit is changing an induced emf

(e)

is set up in the circuit whose magnitude is equal to the negative rate of change of magnetic flux

(ϕ)

e = - \frac{d ϕ}{d t}

Given,

d ϕ = 8 \times 1 0^{- 4} Wb,

d t = 0.4 s

\Rightarrow e = - \frac{8 \times 1 0 ^{- 4}}{0.4}

= - 2 \times 1 0^{- 3} V

Note : The direction of induced emf is such as to oppose the change that produced it.

Q. A small piece of metal wire is dragged across the gap between the poles of a magnet is $0.4 s$ . If the change in magnetic flux in the wire is $8 \times 10^{- 4} Wb$ , then emf induced in the wire is:

$8 \times 10^{- 3} V$

$6 \times 10^{- 3} V$

$4 \times 10^{- 3} V$

$2 \times 10^{- 3} V$

Q. A small piece of metal wire is dragged across the gap between the poles of a magnet is 0.4s. If the change in magnetic flux in the wire is 8×10−4Wb, then emf induced in the wire is:

8×10−3V

6×10−3V

4×10−3V

2×10−3V

Q. A small piece of metal wire is dragged across the gap between the poles of a magnet is $0.4 s$ . If the change in magnetic flux in the wire is $8 \times 10^{- 4} Wb$ , then emf induced in the wire is:

$8 \times 10^{- 3} V$

$6 \times 10^{- 3} V$

$4 \times 10^{- 3} V$

$2 \times 10^{- 3} V$