Question
Consider an infinitely long wire carrying a current I (t), with $\frac{\text{dI}}{\text{dt}}=\lambda=\text{constant}$. Find the current produced in the rectangular loop of wire ABCD if its resistance is R (Fig).
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Answer
To approach these types of problems integration is very useful to find the total magnetic flux linked with the loop.
Let us first consider an elementary strip of length l and width dr at a distance r from an infinite long current carrying wire. The magnetic field at strip due to current carrying wire is given by,
$\vec{\text{B}}\text{(r)}=\frac{\mu_0\text{I}}{2\pi\text{r}}$ (out of paper)
Area of the elementary strip is, dA = l.dr
So, total flux through the loop is
$\phi_\text{m}=\vec{\text{B}}.\vec{\text{A}}=\frac{\mu_0\text{I}}{2\pi}\text{I}\int_{\text{x}_0}^{\text{x}}\frac{\text{dr}}{\text{r}}=\frac{\mu_0\text{Il}}{2\pi}\text{ln}\frac{\text{x}}{\text{x}_0}\ .....(\text{i})$
The emf induced can be obtained by differentiating the eq. (i) w.r.t. t and then applying Ohm's law
$\text{I}=\frac{\in}{\text{R}}$
And $|\in|=\frac{\text{d}\phi}{\text{dt}}$
We have, indueced current
$\text{I}=\frac{1}{\text{R}}\frac{\text{d}\phi}{\text{dt}}=\frac{\mu_0\text{l}}{2\pi}\frac{\lambda}{\text{R}}\text{in}\frac{\text{x}}{\text{x}_0}\ \ \bigg(\because\frac{\text{dI}}{\text{dt}}=\lambda\bigg)$
Let us first consider an elementary strip of length l and width dr at a distance r from an infinite long current carrying wire. The magnetic field at strip due to current carrying wire is given by,
$\vec{\text{B}}\text{(r)}=\frac{\mu_0\text{I}}{2\pi\text{r}}$ (out of paper)
Area of the elementary strip is, dA = l.dr
So, total flux through the loop is
$\phi_\text{m}=\vec{\text{B}}.\vec{\text{A}}=\frac{\mu_0\text{I}}{2\pi}\text{I}\int_{\text{x}_0}^{\text{x}}\frac{\text{dr}}{\text{r}}=\frac{\mu_0\text{Il}}{2\pi}\text{ln}\frac{\text{x}}{\text{x}_0}\ .....(\text{i})$
The emf induced can be obtained by differentiating the eq. (i) w.r.t. t and then applying Ohm's law
$\text{I}=\frac{\in}{\text{R}}$
And $|\in|=\frac{\text{d}\phi}{\text{dt}}$
We have, indueced current
$\text{I}=\frac{1}{\text{R}}\frac{\text{d}\phi}{\text{dt}}=\frac{\mu_0\text{l}}{2\pi}\frac{\lambda}{\text{R}}\text{in}\frac{\text{x}}{\text{x}_0}\ \ \bigg(\because\frac{\text{dI}}{\text{dt}}=\lambda\bigg)$
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