Answer :
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 = ?
Answer:
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.