Question

A magnetic field is aligned perpendicular to the plane of a circular loop of wire with radius 5.0 cm and 20 turns. The magnetic field B changes with time t according to the following function: B 12 exp(-3t) +1.5t +6 T Determine the induced emf in the circular loop of wire as a (i) function of time, and the time at which the induced emf is 0.1 V. (ii) If the magnetic fieldis instead aligned at 30° with respect to the normal of the circular loop of wire, find the induced emf as a function of time.

Answer 1

Answer:

$V = 0.157(-36 e^{-3t} + 1.5)$

$t = 1.24 s$

Part b)

$EMF = 0.136(-36e^{-3t} + 1.5)$

Explanation:

magnetic field due to external source is given as

$B = 12 e^{-3t} + 1.5 t + 6$

area of the loop is given as

$A = \pi r^2$

$A = \pi(0.05)^2$

$A = 7.85 \times 10^{-3} m^2$

Now we have

$\phi = NBAcos0$

$\phi = (20)(12e^{-3t} + 1.5t + 6)(7.85 \times 10^{-3})$

$V = \frac{d\phi}{dt}$

$V = 0.157\frac{d}{dt}(12e^{-3t} + 1.5t + 6)$

$V = 0.157(-36 e^{-3t} + 1.5)$

now we need to find the time at which voltage is 0.1 Volts so we have

$0.1 V = 0.157(-36e^{-3t} + 1.5)$

$t = 1.24 s$

Part b)

If magnetic field is inclined at an angle of 30 degree with the normal of the loop then

$\phi = NBAcos30$

now we know that induced EMF is given as

$EMF = \frac{d\phi}{dt}$

$EMF = NAcos30\frac{dB}{dt}$

$EMF = (20)(7.85 \times 10^{-3})cos30(\frac{d}{dt}(12e^{-3t} + 1.5t + 6))$

$EMF = 0.136(-36e^{-3t} + 1.5)$