A circular coil of radius 5.0 cm and resistance 0.20 Ω is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.50e −20t T. What is the magnitude of the current induced in the coil at the time t = 2.0 s?

Given Information:

Radius of loop = r = 5 cm = 0.05 m

Resistance = R = 0.20 Ω

Magnetic field = B = (0.50e^−20t)  T  

Time = t = 2 sec

Required Information:  

Magnitude of current = I = ?  

Magnitude of current = 4.168x10⁻²¹ A

Explanation:

The magnitude of the current can be calculated using

I = ξ/R

Where ξ is induced emf in the circular coil and is given by

ξ = -Nd/dt(Φ)

Where Φ is the flux and is given by

Φ = BA

Where B is the magnetic field and A is the area of circular coil .

ξ = -Nd/dt(BA )

ξ = -NAd/dt(B )

Magnetic field (B) is given as a function of time 0.50e^−20t taking its derivative with respect to time yields,

d/dt(B )

d/dt(0.50e^−20t)

(0.50/-20)e^−20t

-0.025e^−20t

at time t = 2 seconds

-0.025e^−20(2)

-10.62x10⁻²⁰ T

ξ = -NA(-10.62x10⁻²⁰)

Area is given by

A = πr²

A = π(0.05)²

A = 0.00785 m²

ξ = -N(0.00785)(-10.62x10⁻²⁰)

Assuming N = 1 the induced emf will be

ξ = -1*(0.00785)(-10.62x10⁻²⁰)

ξ = 8.337x10⁻²² V

I = ξ/R

I = 8.337x10⁻²²/0.20

I = 4.168x10⁻²¹ A

Therefore, the magnitude of the current induced in the coil at the time t = 2.0 sec is 4.168x10⁻²¹ A.

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