Answer:
0.0133 A
Explanation:
The time at which B=1.33 T is given by
1.33 = 0.38*t^3
t = (1.33/0.38)^(1/3) = 1.52 s
Using Faraday's Law, we have
emf = - dΦ/dt = - A dB/dt = - A d/dt ( 0.380 t^3 )
Area A = pi * r² = 3.141 *(0.025 *0.025) = 0.00196 m²
emf = - A*(3*0.38)*t^2
thus, the emf at t=1.52 s is
emf = - 0.00196*(3*0.38)*(1.52)^2 = -0.0052 V
if the resistance is 0.390 ohms, then the current is given by
I = V/R = 0.0052/0.390 = 0.0133 A
The current in the loop (magnitude and direction) at the instant when b = 1.33 t is; I = 0.0133 A and flowing in the opposite direction
We are given;
Radius of loop; r = 0.025 m
Resistance; R = 0.390 Ω
b(t) = 0.38t³ T/s³
Now, we are given the instance when B = 1.33 T. Thus, time at this magnitude of magnetic field gives;
1.33 = 0.38t³
t³ = 1.33/0.38
t = ∛3.5
t = 1.52 s
From faraday's law, the EMF is expressed as;
emf = -A(dB/dt)
Now, we have
b(t) = 0.38t³ T/s³
Thus;
dB/dT = 3(0.38t²)
Area is; A = πr²
A = π(0.025²)
A = 0.00196 m²
Plugging the relevant values into the emf equation gives us;
emf = 0.00196 × (3 × 0.38) × (1.52)²
emf = -0.0052 V
Formula for current here is;
I = V/R
Where V is emf or voltage and R is resistance. Thus;
I = -0.0052/0.390
I = -0.0133 A
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