Question

The 10-cm-wide, zero-resistance slide wire shown in FIGURE is pushed toward the $2.0\Omega$resistor at a steady speed of $0.50m/s$. The magnetic field strength is $0.50T$.

a. How big is the pushing force?

b. How much power does the pushing force supply to the wire?

c. What are the direction and magnitude of the induced current?

d. How much power is dissipated in the resistor?

Short answer

(a) Pushing Force, $F=6.25\times {10}^{-4}\mathrm{N}$

(b) Power, $P=3\times {10}^{-4}\mathrm{W}$

(c) Induced current, $I=12.5\mathrm{mA}$counterclockwise

(d) Power dissipated in resistor,$P=3\times {10}^{-4}\mathrm{W}$

Step-by-step solution

Find Pushing force (part a)

When a current-carrying wire moves throughout a magnetic field, the field imposes a magnetic force on the wire. The magnetic force exerted by a wire of length is given by the equation in the form

$F=IlB$

Where $l$is the width of the wire and $I$ is the current in the wire. To determine the magnitude of the emf, we can restate Faraday's law using equation

$\epsilon =Blv$

We use Ohm's law, to get the Induced current

$I=\frac{\epsilon }{R}=\frac{Blv}{R}$

$F=IlB=\left(\frac{Blv}{R}\right)lB=\frac{B^2{l}^{2}v}{R}$

Now, we plug the values for $B,l,v$and $R$into equation to get $F$

$F=\frac{B^2{l}^{2}v}{R}$

$=\frac{(0.50\mathrm{T}{)}^2(0.10\mathrm{m}{)}^2(0.50\mathrm{m}/\mathrm{s})}{2\Omega }$

$=6.25\times {10}^{-4}\mathrm{N}$

Find Power 

The magnetic force allows work to be done on the wire and energy to be supplied to it. The power is expressed by the rate of work done, and the energy due to pulling force is given by equation.

$P=F_{\text{push}}v$

we plug the values for $F_{\text{push}}$and$v$into equtaion to get $P$

$=\left(6.25\times {10}^{-4}\mathrm{N}\right)(0.50\mathrm{m}/\mathrm{s})$

$=3\times {10}^{-4}\mathrm{W}$

Find induced current (part b)

To get the induced current, we plug the values for $B,l,v$and $R$into equation

$I=\frac{Blv}{R}$

$=\frac{(0.50\mathrm{T})(0.10\mathrm{m})(0.50\mathrm{m}/\mathrm{s})}{2\Omega }$

$=12.5\times {10}^{-3}\mathrm{A}$

$=12.5\mathrm{mA}$

The applied magnetic field strength decreases as the wire moves to the right, and according to Lenz's law, when the applied magnetic drops, the induced magnetic field in the wire moves in the same direction as the applied magnetic field, which is out of the page in this case.

The right-hand rule can be used to establish that direction of the induced current; because the magnetic field is outside of the page, your thumb will point counterclockwise.

Find Powe Dissipated in resistor (part c)

We use the value of the induced current and the resistance $R$to get the dissipated power in the wire by

$P=I^{2}R$

$={\left(12.5\times {10}^{-3}\mathrm{A}\right)}^2\left(2\Omega \right)$

$=3\times {10}^{-4}\mathrm{W}$