Question

A circular loop of flexible iron wire has an initial circumference of 165.0cm, but its circumference is decreasing at a constant rate of 12.0cm/s due to a tangential pull on the wire. The loop is in a constant, uniform magnetic field oriented perpendicular to the plane of the loop of magnitude 0.500T.

Answer 1

Answer:

0.005 V

Explanation:

We are given that

Initial circumference of circular loop=C=165 cm

Rate of circumference,$\frac{dC}{dt}=12 cm/s$

Magnetic field,B=0.5 T

We have to find the induced emf at time t=9 s

We know that induced amf,E=$\frac{Bd(A)}{dt}$

Area of circular coil,A=$\pi r^2$

$E=B\frac{d(\pi r^2)}{dt}=B(2\pi r)\frac{dr}{dt}$

Circumference of circular coil,C=$2\pi r$

$165=2\pi r$

$r=\frac{165}{2\pi}$

$\frac{dr}{dt}=\frac{1}{2\pi}\frac{dC}{dt}=\frac{1}{2\pi}\times (12)=\frac{6}{\pi} cm/s=\frac{6\times 10^{-2}}{\pi} m/s$

Radius of coil at time t=9 s

$r=\frac{165}{2\pi}-(\frac{6}{\pi}\times 9)=9.08 cm=9.08\times 10^{-2} m$

$1 m=100 cm$

E=$-0.5(2\pi\times 9.08\times 10^{-2})\times \frac{6\times 10^{-2}}{\pi}=-0.005 V$

Magnitude of induced emf=0.005 V

Answer 2

Answer:

Explanation:

initial circumference, C = 165 cm = 1.65 m

rate of change of circumference, dC/dt = 12 cm /s = 0.12 m/s

magnetic field, B = 0.5 T

According to the Faraday's law of electromagnetic induction

e = dФ/dt

where, Ф is the magnetic flux

Ф = B A

where, A is the area of the coil

$e=\frac{d}{dt}(BA)$

$e=B\frac{dA}{dt}$

$e=B\frac{d(\pi r^{2})}{dt}$

$e=2\pi r\times B\frac{dr}{dt}$     ... (1)

C = 2πr

dC/dt = 2π dr/dt

Put in equation (1)

$e=C \times B\times \frac{1}{2\pi }\times \frac{dC}{dt}$

e = (1.65 x 0.5 x 0.12) / (2 x 3.14)

e = 0.016 V