Faraday's law: A loop of radius r = 3.0 cm is placed parallel to the xy-plane in a uniform magnetic = 0.75 T . The resistance of the loop is 18 Ω. Starting at t = 0, the magnitude of the field decreases uniformly to zero in 0.15 seconds. What is the magnitude of the electric current produced in the loop during that time?
A) 0.79 mA
B) 3.9 mA
C) 1.7 mA
D) 2.1 mA
E) 0.20 mA

Question

Faraday's law: A loop of radius r = 3.0 cm is placed parallel to the xy-plane in a uniform magnetic  = 0.75 T  . The resistance of the loop is 18 Ω. Starting at t = 0, the magnitude of the field decreases uniformly to zero in 0.15 seconds. What is the magnitude of the electric current produced in the loop during that time?
A) 0.79 mA 
B) 3.9 mA 
C) 1.7 mA 
D) 2.1 mA 
E) 0.20 mA

Quizplus · Accepted Answer

The induced EMF in the loop is given by:

$EMF = -\frac{d\phi_B}{dt}$

where $\phi_B$ is the magnetic flux through the loop.

The magnetic flux through the loop is given by:

$\phi_B = BA \cos 	heta$

where $B$ is the magnetic field strength, $A$ is the area of the loop, and $	heta$ is the angle between the normal to the loop and the magnetic field.

Since the loop is parallel to the xy-plane, $	heta = 0$ and $\cos 	heta = 1$. Also, the area of the loop is $A = \pi r^2 = (3.0	ext{ cm})^2 \pi = 28.3 	ext{ cm}^2 = 2.83 	imes 10^{-3} 	ext{ m}^2$.

Therefore, the magnetic flux through the loop is:

$\phi_B = BA = (0.75	ext{ T})(2.83 	imes 10^{-3} 	ext{ m}^2) = 2.12 	imes 10^{-3} 	ext{ Wb}$

The rate of change of the magnetic flux is:

$\frac{d\phi_B}{dt} = \frac{\Delta \phi_B}{\Delta t} = \frac{2.12 	imes 10^{-3} 	ext{ Wb}}{0.15	ext{ s}} = 14.1 	ext{ mWb/s}$

The induced EMF in the loop is therefore:

$EMF = -\frac{d\phi_B}{dt} = -(14.1 	ext{ mWb/s}) = -14.1 	ext{ mV}$

The direction of the induced EMF is given by Lenz's law, which states that the direction of the induced EMF is such that it opposes the change in magnetic flux. Since the magnetic field is decreasing, the induced EMF will produce a current that creates a magnetic field that opposes the decrease in the original magnetic field.

The magnitude of the induced current is:

$I = \frac{EMF}{R} = \frac{-14.1 	ext{ mV}}{18 \Omega} = -0.78	ext{ mA}$

Taking the absolute value, the magnitude of the induced current is:

$I = 0.78	ext{ mA} \approx \boxed{	extbf{(A) }0.79 	ext{ mA}}$

Faraday's Law: a Loop of Radius R = 3