Chapter 21, Problem 14P

To determine

The force required to pull the loop from the field.

Answer to Problem 14P

Solution:

0.499 N.

Explanation of Solution

Given:

The rectangular loop is placed in a uniform magnetic field of 0.550 T.

The total resistance of the loop is $0.230\Omega$ .

The constant velocity required to pull the loop is $3.10\text{ m/s}$ .

The dimension of the rectangular loop is $0.350\text{ m}\times \text{0}\text{.750 m}$ .

Formula Used:

The magnitude of the induced EMF is as follows:

  $\epsilon =Blv$

Here, B is the magnetic field, l is the length, and v is the velocity.

The expression for the induced Current is as follows:

  $I=\frac{\epsilon }{R}$

Here, $R$ is the resistance.

The expression for the force is as follows:

  $F=IBl$

Here, I is the current.

Calculate:

According to Lenz’s law, the induced current is produced in the loop due to the change in the magnetic flux. The changing magnetic flux produces an induced magnetic field, which produces an induced EMF.

As the loop is pulled away from the uniform magnetic field, an induced EMF is produced.

From Lenz’s law, the induced EMF is directed into the page due to the decreasing inward magnetic flux through the loop and this results in a clockwise induced current produced in the loop.

Due to the induced current on the left side of the loop, there will be a magnetic force acting on it directed to the left. This opposing force must be equal to the force directed to the right to keep the rod moving with constant velocity.

Substitute $\frac{\epsilon }{R}$ for I and $Blv$ for $\epsilon$ in the equation $F=IBl$ .

  $\begin{array}{c}F=\left(\frac{\epsilon }{R}\right)Bl\end{array}$

   $\begin{array}{c}=\frac{B^2{l}^{2}v}{R}\end{array}$

Therefore, the force is as follows:

  $F=\frac{B^2{l}^{2}v}{R}$

Now, substitute 0.550 T for B, 3.10 m/s for v , $0.230\Omega$ for R , and 0.350 m for l in the equation $F=\frac{B^2{l}^{2}v}{R}$ .

  $\begin{array}{c}F=\frac{{\left(0.550\text{ T}\right)}^2{\left(0.350\text{ m}\right)}^2\left(3.10\text{ m/s}\right)}{0.230\Omega }\end{array}$

   $\begin{array}{c}=0.499\text{ N}\end{array}$

Conclusion:

Hence, the force required to pull the loop from the field is 0.499 N.