Question

A $3.0-cm$-diameter, $10$-turn coil of wire, located at $z=0$in the$xy$-plane, carries a current of $2.5A.A2.0-mm$-diameter conducting loop with $2.0*10-4\Omega$resistance is also in the $xy$plane at the center of the coil. At $t=0s$, the loop begins to move along the $z$-axis with a constant speed of $75m/s$. What is the induced current in the conducting loop at$t=200ms?$The diameter of the conducting loop is much smaller than that of the coil, so you can assume that the magnetic field through the loop is everywhere the on-axis field of the coil.

Short answer

On-axis field of the coil,$I_l=45\mathrm{mA}$

Step-by-step solution

Faraday' principle

The principle of Faraday,

According to Faraday's equation of induction, the amplitude of the electromotive force in a circuits is proportional to the rate of rate of change of magnetic that cuts across the circuit.

Simplification

Given values:

$r_c=1.5\mathrm{cm}$

$r_l=1.0\mathrm{mm}$

$R_l=2.0\times {10}^{-4}\Omega$

$I_c=2.5\mathrm{A}$

$v=75\mathrm{m}/\mathrm{s}$

Using the formula,

$B_c=N\frac{{\mu }_o}{2}\frac{I_c{r}_c^2}{{\left(z^2+r_c^2\right)}^{3/2}}$

$\frac{d{B}_c}{dz}=N\frac{{\mu }_o}{2}\left(-\frac{3}{2}\right)\frac{I_c{r}_c^2}{{\left(z^2+r_c^2\right)}^{5/2}}\left(2z\right)$

$\frac{d{B}_c}{dz}=-\frac{3}{2}N{\mu }_o\frac{I_c{r}_c^2}{{\left(z^2+r_c^2\right)}^{5/2}}z$

From Faraday's law,

${\epsilon }_l=\left|\frac{d{\Phi }_m}{dt}\right|$

${\epsilon }_l=\frac{d\left(A_{l}B\cos \theta \right)}{dt}$

${\epsilon }_l=A_l\cos \theta \frac{dB}{dt}$

${\epsilon }_l=A_l\cos \theta \frac{dB}{dt}\frac{dz}{dt}$

Find the value of Il

We know that, $\epsilon =IR$:

${\epsilon }_l=I_l{R}_l$

$I_l=\frac{A_l}{R_l}\cos \theta \left(\frac{d{B}_c}{dz}\right)\left(\frac{d{z}_l}{dt}\right)$

$I_l=\frac{\pi r_l^2}{R_l}\cos \theta \left(\frac{3}{2}N{\mu }_o\frac{I_c{r}_c^2}{{\left(z^2+r_l^2\right)}^{5/2}}z\right)\left(\frac{d{z}_l}{dt}\right)$

$I_l=\frac{\pi (1\mathrm{mm}{)}^2}{\left(2.0\times {10}^{-4}\Omega \right)}\left(\frac{3}{2}\left(10\right)\left(1.26\times {10}^{-6}\right)\frac{(2.5\mathrm{A})(1.5\mathrm{m}{)}^2}{{\left((0.015\mathrm{m}{)}^2+(1.5\mathrm{cm}{)}^2\right)}^{5/2}}(0.015\mathrm{m})\right)(75\mathrm{m}/\mathrm{s})$

$I_l=45\mathrm{mA}$