A small piece of metal wire is dragged across the gap between the pole pieces of a magnet in 0.5 second. The magnetic flux between the pole pieces is known to be 8 × 10-4 Wb. The emf induced in the wire is

Question

A small piece of metal wire is dragged across the gap between the pole pieces of a magnet in 0.5 second. The magnetic flux between the pole pieces is known to be 8 × 10-4 Wb. The emf induced in the wire is (A) 16mV (B) 1.6V (C) 1.6mV (D) 16V

Tardigrade Admin · Accepted Answer

Induced emf e=(-dφ/dt).Assuming, small change in flux dφ=8Ã10-4Wb change in time dt== 0.5s |e|=(8Ã10-4/0.5)=(80Ã10-4/5) =16Ã10-4=1.6Ã10-3V=1.6mV

Q. A small piece of metal wire is dragged across the gap between the pole pieces of a magnet in 0.5 second. The magnetic flux between the pole pieces is known to be 8 × $1 0^{- 4}$ Wb. The emf induced in the wire is

16mV

1.6V

1.6mV

16V

Induced emf e= $\frac{- d ϕ}{d t} .$ Assuming, small
change in flux $d ϕ = 8 \times 1 0^{- 4}$ Wb
change in time dt== 0.5s
$∣ e ∣ = \frac{8 \times 1 0 ^{- 4}}{0.5} = \frac{80 \times 1 0 ^{- 4}}{5}$
=16× $1 0^{- 4} = 1.6 \times 1 0^{- 3} V = 1.6 mV$

Q. A small piece of metal wire is dragged across the gap between the pole pieces of a magnet in 0.5 second. The magnetic flux between the pole pieces is known to be 8 × 10−4 Wb. The emf induced in the wire is

16mV

1.6V

1.6mV

16V

Induced emf e=dt−dϕ​.Assuming, small change in flux dϕ=8×10−4Wb change in time dt== 0.5s ∣e∣=0.58×10−4​=580×10−4​ =16×10−4=1.6×10−3V=1.6mV

Q. A small piece of metal wire is dragged across the gap between the pole pieces of a magnet in 0.5 second. The magnetic flux between the pole pieces is known to be 8 × $1 0^{- 4}$ Wb. The emf induced in the wire is

Induced emf e= $\frac{- d ϕ}{d t} .$ Assuming, small
change in flux $d ϕ = 8 \times 1 0^{- 4}$ Wb
change in time dt== 0.5s
$∣ e ∣ = \frac{8 \times 1 0 ^{- 4}}{0.5} = \frac{80 \times 1 0 ^{- 4}}{5}$
=16× $1 0^{- 4} = 1.6 \times 1 0^{- 3} V = 1.6 mV$