Question

A coil 3.95 cm radius, containing 520 turns, is placed in a uniform magnetic field that varies with time according to B=( 1.20×10−2 T/s )t+( 3.45×10−5 T/s4 )t4. The coil is connected to a 560-Ω resistor, and its plane is perpendicular to the magnetic field. You can ignore the resistance of the coil.
a-)Find the magnitude of the induced emf in the coil as a function of time.

Find the magnitude of the induced emf in the coil as a function of time.

E= 1.07×10−2 V +( 1.05×10−4 V/s3 )t3
E= 3.36×10−2 V +( 8.26×10−5 V/s3 )t3
E= 3.36×10−2 V +( 3.30×10−4 V/s3 )t3
E= 1.07×10−2 V +( 3.30×10−4 V/s3 )t3
b-)What is the current in the resistor at time t0 = 5.25 s ?

Answer 1

Answer:

(a) E= 3.36×10−2 V +( 3.30×10−4 V/s3 )t3

(b) $I=0.0085\ A$

Explanation:

Given:

• radius if the coil, $r=0.0395\ m$

• no. of turns in the coil, $n=520$

• variation of the magnetic field in the coil, $B=(1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4$

• resistor connected to the coil, $R=560\ \Omega$

(a)

we know, according to Faraday's Law:

$emf=n.\frac{d\phi}{dt}$

where:

$d \phi=$ change in associated magnetic flux

$\phi= B.A$

where:

A= area enclosed by the coil

Here

$A=\pi.r^2$

$A=\pi\times 0.0395^2$

$A=0.0049\ m^2$

$\therefore \phi=((1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4)\times 0.0049$

So, emf:

$emf= 520\times \frac{d}{dt} [((1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4)\times 0.0049]$

$emf= 520\times 0.0049\times \frac{d}{dt} [(1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4)]$

$emf= 2.548\times [0.012+(13.8\times 10^{-5})t^3)]$

$emf= 0.0306+3.516\times 10^{-4}\ t^3$

(b)

Given:

$t_0=5.25\ s$

Now, emf at given time:

$emf=4.7755\times 10^{-2}\ V$

∴Current

$I=\frac{emf}{R}$

$I=\frac{4.7755\times 10^{-2}}{560}$

$I=8.5\times 10^{-5} A$