91Ó°ÊÓ

We have a rectangular loop of wire with dimensions w = 1.50 cm by L = 8.00 cm and resistance R = 0.600$\mathrm{Î\copyright }$is being pulled to the right out of a region of a uniform magnetic field (with a magnitude of B = 2.40 T) with a speed of v = 3.00 m/s. Let x be the portion of the side that is still in the magnetic field as shown in the following figure. The magnetic flux is, therefore,

${\mathrm{Ï•}}_B=BA=Bwx$

The magnitude of the induced magnetic field is, therefore,

$\mathrm{Î\mu }=\frac{d{\mathrm{Ï•}}_B}{dt}=Bw\frac{dx}{dt}=Bwv$

The induced current is, therefore,

$I=\frac{\mathrm{Î\mu }}{R}=\frac{Bwv}{R}$

The force that acts on the loop due to the external magnetic field is only the force on the left side (that inside the magnetic field), that is,

$F_B=IwB$

Substitute with I, we get,

$F_B=\frac{B^2{w}^{2}v}{R}$

Substitute with the givens we get,

role="math" localid="1664177621732" $$\begin{gathered} F_B=\frac{(2.40T{)}^2(0.0150m{)}^2(3.00m/s)}{0.600\mathrm{Î\copyright }}=6.48×{10}^{-3}N \\ {F}_B=6.48×{10}^{-3}N \end{gathered}$$

The flux is decreasing, so the loop is pulled toward the magnetic field to increase the flux through the loop, thus the force points toward the left.