Question

A coil 3.55 cm in 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+( 2.85×10^−5 T/s4 )t^4. The coil is connected to a 610 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 t0 = 5.20 s ?

Answer 1

The Electric current is 1.11* 10^{-4}A


Given that the coil's radius is 3.55 cm (0.35 m),

The formula for the coil's area is A = r2 A = (3.14) (0.35)2 = 0.005024 m2.

R = Resistance = 600 N = Number of spins = 500 B = Magnetic field = (0.0120)

t + (3 x 10⁻⁵) t⁴

The number t = 5 is substituted for taking the derivative at both the induced current and the electric current.

The Electric current is therefore 1.11* 10^{-4}A
Electric current - The rate of electron passage in a conductor is known as electric current. The ampere is the electric current's SI unit. Electrons are little particles that are part of a substance's molecular structure. These electrons can be held loosely or securely depending on the situation.

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Answer 2

The current is 1.13* 10^{-4}A

Given that r = radius of the coil = 4 cm = 0.04 m

Area of coil is given as

A = πr²

A = (3.14) (0.04)² = 0.005024 m²

N = Number of turns = 500

R = Resistance = 600 Ω

B = magnetic field = (0.0120)t + (3 x 10⁻⁵) t⁴

Taking derivative at both the side

$\frac{dB}{dt} = (0.120 + (12 \times 10^-5)t^3)$

Induced current is given as

$i= (\frac{NA}{R} )( \frac{db}{dt} )$

$i \: = (\frac{NA}{R})(12 \times 10 ^{-5} )t^3$

substituting the value t = 5

$i = ( \frac{(500)(0.005024)}{600}) (12 \times 10 ^{-5} )5^3$

$i = 1.13 \times 10 ^{ - 4} A$

Hence the current is

$1.13 \times 10^-4A$

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