A coil 4.40 cm radius, containing 470 turns, is placed in a uniform magnetic field that varies with time according to B=( 1.20×10−2 T/s )t+( 2.90×10−5 T/s4 )t4. The coil is connected to a 530-Ω resistor, and its plane is perpendicular to the magnetic field. You can ignore the resistance of the coil.
Part A
Find the magnitude of the induced emf in the coil as a function of time.
E= 1.09×10−2 V +( 1.06×10−4 V/s3 )t3
E= 3.43×10−2 V +( 8.29×10−5 V/s3 )t3
E= 3.43×10−2 V +( 3.32×10−4 V/s3 )t3
E= 1.09×10−2 V +( 3.32×10−4 V/s3 )t3
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Part B
What is the current in the resistor at time t0 = 4.50 s ?
I =

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AnnuG AnnuG · Answer 1 · 2022-12-03T18:40:01+01:00

The current in the resistor at time 4.50 s is 1.2179 X 10⁻⁴ A.

Number of turns of coil = N = 470

Radius of the coil = R = 4.40 cm = 0.044 m

Magnetic field varying according to time = B = ( 1.20×10⁻² T/s )t +(2.90×10⁻⁵ T/s⁴ )t⁴

Resistance of the resistor = 530 Ω

Part A :

The flux over the coil is = φ =

= φ = nBA

= φ = 470 X B X (π X 0.044)

= φ = 2.85 (( 1.20 × 10⁻² T/s )t +(2.90×10⁻⁵ T/s⁴ )t⁴)

The magnitude of EMF is = E =

= E = dφ / dt

= E = 2.85 (1.20 × 10⁻² + 11.60 × 10⁻⁵ t³)

= E = (3.43 × 10⁻² V ) + ( 3.32 × 10⁻⁴ V/s³) t³

Part B:

The current in the coil = I =

= I = E / R

= I = (3.43 × 10⁻² V ) + ( 3.32 × 10⁻⁴ V/s³) t³ / 530

= I = 1.2179 X 10⁻⁴ A

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