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

Answer:

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

$B=1.2\times 10^{-2}t+2.6\times 10^{-5}t^{4}$

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

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

magnetic flux, Ф = N x B x A x Cos 0

Ф = N A B

Differentiate both sides

$\frac{d\phi}{dt}=NA\frac{dB}{dt}$

$\frac{d\phi}{dt}=500\times 3.14 \times 0.042\times 0.042\times \left ( 1.2\times 10^{-2}+4 \times 2.6\times 10^{-5}t^{3}\right )$

So, the magnitude of induced emf is given by

$e =3.324\times 10^{-2}+28.8 \times 10^{-5}t^{3} V$

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

Answer 2

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 :

$B=1.2\times 10^{-2}t+2.6\times 10^{-5}t^4$

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 :

$\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=\dfrac{-d(NBA)}{dt}\\\\\epsilon=N\pi r^2\dfrac{-dB}{dt}\\\\\epsilon=N\pi r^2\times \dfrac{-d(1.2\times 10^{-2}t+2.6\times 10^{-5}t^4)}{dt}\\\\\epsilon=N\pi r^2\times (1.2\times 10^{-2}+10.4\times 10^{-5}t^3)\\\\\epsilon=500\pi \times (4.2\times 10^{-2})^2\times (1.2\times 10^{-2}+10.4\times 10^{-5}t^3)\\\\\epsilon=2.77(1.2\times 10^{-2}+10.4\times 10^{-5}t^3)\ V$

Hence, this is the required solution.