Question

A circular loop of wire with radius r= 0.0480 m and resistance R= 0.160 is in a region of a spatially uniform magnetic field, as shown in Fig. The magnetic field is directed out of the plane of the figure. The magnetic field has an initial value of 8.00 T and is decreasing at a rate of dB/dt= -0.680 T/s.

(a) Is the induced current in the loop clockwise or counterclockwise?

(b) What is the rate at which electrical energy is being dissipated by the resistance of the loop?

Short answer

1. The induced current in the loop is in the counterclockwise direction.
2. The rate at which electrical energy is dissipated by the resistance of the loop is 0.151W.

Step-by-step solution

Concept

The answer to part (a) is given by Lenz's law, which states that the induced current always tends to oppose or cancel out the change that caused it. Our target variable in part (b) is the rate at which energy in the loop is dissipated; Energy is dissipated by the resistance R of the loop at the rate $P_{dissipated}=I^{2}R$. The current I in the loop equal /R, where e is the induced emf is given by Faraday's law.

Calculate the direction of the inducing current.

By Lenz's law, the induced current must produce a magnetic field ${\overrightarrow{B}}_{induced}$inside the loop that is upward, opposing the change in the flux. By using the right-hand rule for the direction of the magnetic field produced by a circular loop,${\overrightarrow{B}}_{induced}$will be in the desired direction if the induced current flows as shown in Fig. given below in the counterclockwise direction.

Calculate electric energy.

The magnetic field is uniform over the loop and is perpendicular to the plane of the loop, so the magnetic flux is ${\varphi }_B=BA$, where $A=\pi r^2$is the area of the circular loop. Hence, by Faraday's law) the induced emf is

$\epsilon =-\frac{d{\varphi }_B}{dt}=-\frac{d}{dt}\left(BA\right)=-A\frac{dB}{dt}=-\pi r^2\frac{dB}{dt}$

So the current in the loop is,

$I=\frac{\left|\epsilon \right|}{R}=\frac{\pi r^2}{R}\left|\frac{dB}{dt}\right|$

Hence,

$P_{dissipated}=I^{2}R={\left(\frac{\pi r^2}{R}\frac{dB}{dt}\right)}^{2}R=\frac{{\pi }^2{r}^4}{R}R{\left|\frac{dB}{dt}\right|}^2$

Substituting the known values of r, dB/dt, and R, we find

$P_{dissipated}=\frac{\pi (0.0480m{)}^4}{0.160\Omega }{\left|0.680T/s\right|}^2=0.151mW$

Where we used$(m^2.T{)}^2/s^2=(Wb/s{)}^2=V^2$ for the units.

The electrical energy keeps being dissipated in the loop at the rate calculated in Step 3 until the magnetic flux (or the magnetic field $\overrightarrow{B}$) stops changing; indeed, substituting dB/dt = 0 into the expression for $P_{dissipated}$gives zero.