Question

A coil 3.85 cm radius, containing 450 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 700-Ω resistor, and its plane is perpendicular to the magnetic field. You can ignore the resistance of the coil.

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

Answer 1

The magnitude of the induced emf is
`E=0.02511V+(0.000217(V)/(s^(3) ) )t^(3)`

Step-by-step explanation:

The area is:

`A=\pi r^(2)`

r = 3.85 cm = 0.0385 m

`A=\pi *0.0385^(2) =0.00465m^(2)`

The induce emf is:

`E=NA(dB)/(dt)`

Replacing

`E=450*0.00465*(((d(1.2x10^(-2))t )/(dt) )+((d(2.6x10^(-5))t^(4) )/(dt) )\\E=2.0925*(1.2x10^(-2) +0.000104t^(3) )\\E=0.02511V+(0.000217(V)/(s^(3) ) )t^(3)`

Answer 2

0.025V + (0.000218V/s³) t³

Step-by-step explanation:

Parameters given:

Radius of coil, r = 3.85 cm = 0.0385 m

Number of turns, N = 450

Magnetic field, B = ( 1.20×10^(−2) T/s )t + (2.60×10^(−5) T/s4 )t^4.

The magnitude of Induced EMF is given as:

E = N * A * dB/dt

Where A is the area of the coil

First, we differentiate the magnetic field with respect to time:

dB/dt = 0.012 + 0.000104t³

Therefore, EMF will be:

E = 450 * 3.142 * (0.012 + 0.000104t³)

E = 2.096(0.012 + 0.000104t³)

E = 0.025V + (0.000218V/s³)t³