Question

A coil 4.00 cm in radius, containing 500 turns, is placed in a uniform magnetic field that varies with time according to B=(0.0120T/s)t+(3.00×10−5T/s4)t4. The coil is connected to a 600−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. (b) What is the current in the resistor at time t = 5.00 s?

Answer 1

The magnitude of the induced emf in the coil as a function of time can be found by calculating the change in magnetic flux through the coil. The current in the resistor at time t = 5.00 s can be calculated using Ohm's Law.

Step-by-step explanation:

In order to find the magnitude of the induced emf in the coil as a function of time, we first need to determine the rate of change of magnetic flux through the coil. The change in magnetic field with respect to time is given by dB/dt=(0.0120 T/s)+(3.00×10⁻⁵ T/s⁴)t³. Since the magnetic field is varying with time, the change in magnetic flux through the coil is given by dΦ=dB/dt * A = [(0.0120 T/s)+(3.00×10⁻⁵ T/s⁴)t³] * π(0.04 m)² = (0.00194 T/s)+(0.000819 T/s⁴)t³. The induced emf in the coil is equal to the negative rate of change of magnetic flux through the coil, so dΦ/dt=-(0.00194 T/s)-(0.000819 T/s⁴)t³.

To find the current in the resistor at time t = 5.00 s, we need to use Ohm's Law, which states that I=V/R (where I is the current, V is the potential difference across the resistor, and R is the resistance). We can find the potential difference by multiplying the induced emf by the number of turns in the coil, so V=-ΔΦ*N=-((0.00194 T/s)-(0.000819 T/s⁴)*5³)*500=-6.325 V (since the induced emf is negative).

Now we can substitute the values into Ohm's Law to find the current: I=V/R=(-6.325 V)/(600 Ω)=-0.01054 A.

Answer 2

A. e = 0.0302 V+ (3.02 x 10-4 V/s)t3

B. I = 1.13 x 10-4 A

Step-by-step explanation:

(a) Consider a coil with radius of r = 4.00 cm, and containing N = 500

turns, it is placed perpendicularly in a magnetic field of B, which is given

by,

B = (0.0120 T/s)t + (3.00 x 10-5 T/s) 14

the magnitude of the induced emf is,

|E 1 =

NdФВ

at

where

B = B A cos(0) = BA. Since the area is constant, then,

|E| = NA (B)

= NA ((0.0120 T/s)t + (3.00 x 10-6 T/s*) 4]

NA[(0.0120 T/s)+(1920 x 10-4 T/s*) t1

(500) (0.040)? [(0.0120 T/s)+ (1.20 x 10-4 T/s^) tº]

= 0.0302 V + (3.02 x 10-4 V/s°) 43

e = 0.0302 V+ (3.02 x 10-4 V/s) t3

(b) The induced emf at t = 5.00 s, is,

e = 0.0302 V +(3.02 x 10-4 V/s) (5.00 s)3

= 0.0680 V

if the resistance of the coil is R = 600 2, then the induced current at t = 5.00

s, is,

E 0.0680 V - 1.13 x 10-4 A

I= = 600 12

I = 1.13 x 10-4 A