Example 29.6 discusses the external force that must be applied to the slide wire to move it at a constant speed. If there were a break in the left-hand end of theU-shaped conductor, how much force would be needed to move the slide wire at constant speed? As in the example, you can ignore friction.

The entire magnetic field that travels across a particular region is measured by magnetic flux.

Magnetic flux is given by

$\mathbf{\varphi }\mathbf{=}\overrightarrow{\mathbf{B}}\mathbf{.}\overrightarrow{\mathbf{A}}$

Where,$\overrightarrow{\mathrm{B}}$ is the magnetic field and$\overrightarrow{\mathrm{A}}$ is the area vector.

Faraday’s law states that, the induced emf in a coil is equal to the negative of the rate of change of magnetic flux times the number of turns in the coil. It involves the interaction of charge with the magnetic field.

$\mathbf{emf}\mathbf{=}\mathbf{-}\mathbf{N}\frac{\mathbf{∆}\mathbf{\varphi }}{\mathbf{∆}\mathbf{t}}$

Here, is the induced emf, N is the number of turns in the coil, and$\varphi$ is the magnetic flux.

In Example 29.6, the applied force is to oppose the magnetic force on the bar due to the current flowing in the bar in the magnetic field. And the current is induced due to the change in the area of the loop resulting in the change in the magnetic flux, according to Faraday's law. But, if there is a break in the loop, there is no induced current since the loop is open$\left(\mathrm{R}=\infty \right)$ .

Thus, there is also no magnetic force, because the bar, in this case, is just a bar with no current in a magnetic field. Therefore, you only need the force that sets the bar in motion, and after that, you do not need any force. Of course, this is because friction is ignored.

Hence, only need the force that sets the bar in motion and after that do not need any force.