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Question

(A) The magnetic field $B$ at a distance $x$ from a long current-carrying wire is given by $B = \frac{{\mu _0}I}{2\pi x}$.Consider a small element of

Vedclass · Accepted Answer

$\frac{{\mu _0}I\upsilon}{{2\pi }} \ln \left( {\frac{{r + l}}{r}} \right)$

Vedclass · Answer

(A) The magnetic field $B$ at a distance $x$ from a long current-carrying wire is given by $B = \frac{{\mu _0}I}{2\pi x}$.Consider a small element of length $dx$ on the rod at a distance $x$ from the wire.The motional emf $d\varepsilon$ induced in this small element is $d\varepsilon = Bv dx = \left( \frac{{\mu _0}I}{2\pi x} \right) \upsilon dx$.To find the total induced emf $\varepsilon$,we integrate this expression from $x = r$ to $x = r + l$:$\varepsilon = \int_{r}^{r+l} \frac{{\mu _0}I\upsilon}{2\pi x} dx$$\varepsilon = \frac{{\mu _0}I\upsilon}{2\pi} \int_{r}^{r+l} \frac{1}{x} dx$$\varepsilon = \frac{{\mu _0}I\upsilon}{2\pi} [\ln x]_{r}^{r+l}$$\varepsilon = \frac{{\mu _0}I\upsilon}{2\pi} \ln \left( \frac{r+l}{r} \right)$.

$A$ conducting rod of length $l$ moves with velocity $\upsilon$ in a direction parallel to a long wire carrying a steady current $I$. The axis of the rod is maintained perpendicular to the wire with the near end at a distance $r$ away as shown in the figure. Find the emf induced in the rod.

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