Question

Question: A thin rectangular coil lies flat on a low-friction table (Figure 22.75). A very long straight wire also lies flat on the table, a distance zfrom the coil. The wire carries a conventional current lto the right as shown, and this current is decreasing: $\mathit{I}\mathbf{}\mathbf{=}\mathbf{}\mathit{a}\mathbf{}\mathbf{-}\mathbf{}\mathit{b}\mathit{t}$, where tis the time in seconds, and aand bare positive constants. The coil has length Land width w, where w << z. It has Nturns of wire with total resistance R.

What are the initial magnitude and direction of the nonzero net force that is acting on the coil?You can neglect friction. Explain in detail. If you must make simplifying assumptions, state clearly what they are, but bear in mind that the net force is not zero.

Short answer

$F_{net}=\frac{{{\mu }_0}^{2}Nab{L}^2}{4{\pi }^{2}R}\left(\frac{1}{z}-\frac{1}{z+w}\right)\ln \left(\frac{z+w}{z}\right)$

Step-by-step solution

Explain the given information

Consider a thin rectangular coil that lie flat on a low friction table. On the same table consider that a very long straight wire is placed at a distance z from the coil. The wire carries a conventional current l that is decreasing $I=a-bt$ in which t represents the time and the a and b are positive constants. Consider the coil has length L and width w, where w << z and it has N turns of wire with resistance R.

Provide the formula

Formula of induced emf

$d\in =N\frac{d\Phi }{dt}\dots \dots \left(1\right)$

Step 3: What are the initial magnitude and direction of the nonzero net force that is acting on the coil?

Consider the coordinate system $xy-\text{plane}$on which the table lies with a long wire parallel to x-axis. According to Lenz’s law, the emf and the current in the coil is in clockwise direction.

Divide the rectangular coil according to the sides like top, bottom, right, and left. The induced current at the top of the coil and the current in the long wire have the same direction. Hence, the net force is,

$F_{net}=F_{top}-F_{bottom}$ ……(2)

Find the induced emf of the coil from Equation (1),

$$\begin{gathered} d\in =N\frac{d\Phi }{dt} \\ d\in =N\frac{d\left(\left(B·L\right)dy\right)}{dt} \end{gathered}$$

To find $\in$integrate the right hand side of the above equation.

$$\begin{gathered} \in =N\int d\in \\ \in =NL{\int }_{y=z}^{z+w}\frac{d}{dt}\left(\frac{{\mu }_{0}I}{2\pi y}dy\right) \\ \in =NL\frac{{\mu }_0}{2\pi }\left({\int }_{y=z}^{z+w}\frac{d\left(a-bt\right)}{dt}\frac{dy}{y}\right) \end{gathered}$$

Find the modulus on both sides,

$$\begin{gathered} \left|\in \right|=\left|\frac{{\mu }_{0}NL}{2\pi }\left(-b{\int }_{y=z}^{z+w}\frac{dy}{y}\right)\right| \\ \in =\frac{{\mu }_{0}NLb}{2\pi }\ln \left(\frac{z+w}{z}\right) \end{gathered}$$

Knowing that,

$I_{induced}=I_{top}=I_{bottom}=\frac{\in }{R}$

$I_{induced}=I_{top}=I_{bottom}=\frac{m{\mu }_{0}BLb}{2\pi R}\ln \left(\frac{z+w}{z}\right)$ ……(3)

Hence,

$$\begin{gathered} F_{net}=\frac{{\mu }_{0}I{I}_{top}L}{2\pi z}-\frac{{\mu }_{0}I{I}_{bottom}L}{2\pi \left(z+w\right)} \\ {F}_{net}=\frac{{\mu }_{0}I{I}_{induced}L}{2\pi }\left(\frac{1}{z}-\frac{1}{z+w}\right) \end{gathered}$$

Substitute Equation (3) in $F_{net}$

$F_{net}=\frac{{{\mu }_0}^{2}NIb{L}^2}{4{\pi }^{2}R}\left(\frac{1}{z}-\frac{1}{z+w}\right)\ln \left(\frac{z+w}{z}\right)$

At t = 0,

$$\begin{gathered} I=a-b\left(0\right) \\ I=a \end{gathered}$$

Therefore,

$F_{net}=\frac{{{\mu }_0}^{2}Nab{L}^2}{4{\pi }^{2}R}\left(\frac{1}{z}-\frac{1}{z+w}\right)\ln \left(\frac{z+w}{z}\right)$

Therefore, the initial magnitude and direction of the nonzero net force that is acting on the coil has been obtained.