Question

A coil 4.20 cm radius, containing 500 turns, is placed in a uniform magnetic field that varies with time according to B=( 1.20×10−2 T/s )t+( 2.60×10−5 T/s4 )t4. The coil is connected to a 640-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.

Answer 1

Step-by-step explanation:

Radius of the coil, r = 4.2 cm

number of turns, N = 500

resistance in the circuit, R = 640 ohm

The magnetic field is given by

`B=1.2* 10^(-2)t+2.6* 10^(-5)t^(4)`

(a) According to the Faraday's law of electromagnetic induction, the magnitude of induced emf is given by

`e = (d\phi)/(dt)`

magnetic flux, Ф = N x B x A x Cos 0

Ф = N A B

Differentiate both sides

`(d\phi)/(dt)=NA(dB)/(dt)`

`(d\phi)/(dt)=500* 3.14 * 0.042* 0.042* \left ( 1.2* 10^(-2)+4 * 2.6* 10^(-5)t^(3)\right )`

So, the magnitude of induced emf is given by

`e =3.324* 10^(-2)+28.8 * 10^(-5)t^(3) V`

This is the magnitude of induced emf as the function of time.

Answer 2

Step-by-step explanation:

Given that,

Radius of the coil, r = 4.2 cm

Number of turns in the coil, N = 500

The magnetic field as a function of time is given by :

`B=1.2* 10^(-2)t+2.6* 10^(-5)t^4`

Resistance of the coil, R = 640 ohms

We need to find the magnitude of induced emf in the coil as a function of time. It is given by :

`\epsilon=(-d\phi)/(dt)\\\\\epsilon=(-d(NBA))/(dt)\\\\\epsilon=N\pi r^2(-dB)/(dt)\\\\\epsilon=N\pi r^2* (-d(1.2* 10^(-2)t+2.6* 10^(-5)t^4))/(dt)\\\\\epsilon=N\pi r^2* (1.2* 10^(-2)+10.4* 10^(-5)t^3)\\\\\epsilon=500\pi * (4.2* 10^(-2))^2* (1.2* 10^(-2)+10.4* 10^(-5)t^3)\\\\\epsilon=2.77(1.2* 10^(-2)+10.4* 10^(-5)t^3)\ V`

Hence, this is the required solution.