A coil 4.50 cm radius, containing 490 turns, is placed in a uniform magnetic fie

Question

A coil 4.50 cm radius, containing 490 turns, is placed in a uniform magnetic field that varies with time according to B=( 1.20×102 T/s )t+( 3.30×105 T/s4 )t4. The coil is connected to a 550- 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.

C

Correct

Part B

What is the current in the resistor at time t0 = 4.80 s ?

E= 1.19×102 V +( 1.31×104 V/s3 )t3 E= 3.74×102 V +( 1.03×104 V/s3 )t3 E= 3.74×102 V +( 4.11×104 V/s3 )t3 E= 1.19×102 V +( 4.11×104 V/s3 )t3

Explanation / Answer

emf = - N*d(flux) / dt

flux = B*area

And, you need to derivative the flux relative to time, So let's do it like that :

B =( 1.20×102 T/s )t+( 3.30×105 T/s4 )t4

area = pi*(4.5/100)^2

flux = ( 1.20×102 T/s )t+( 3.30×105 T/s4 )t4 * pi*(4.5 / 100)^2

then :

emf = -490*d( 1.20×102 T/s )t+( 3.30×105 T/s4 )t4 *pi*(4.5 / 100)^2) / dt

emf = -3.12*d( 1.20×102 T/s )t+( 3.30×105 T/s4 )t4/ dt

emf = -3.12*( 1.20×102 T/s )+4* ( 3.30×105 T/s3 )t3

emf = - 3.74 x 102 V + (4.11 x 104 V/s3)t3

therefore matches with option c.

for 4.8 s:

emf = - (3.74 x 102 V + (4.11 x 104 V/s3)t3)

emf = - 8.28 x 10-2 V

current = emf/R = (8.28 x 10-2)/550 = 150.5 x 10-6 Ampere =150.5 micro-ampere

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