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(B) The magnetic field $B$ at a distance $r = a + x$ from the wire is given by $B = \frac{\mu_0 i}{2 \pi (a + x)}$.Consider a small element of length

Vedclass · Accepted Answer

$\frac{{{\mu _0}i\omega }}{{2\pi }}\left[ {l - a\ln \left( {\frac{{l + a}}{a}} \right)} \right]$

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(B) The magnetic field $B$ at a distance $r = a + x$ from the wire is given by $B = \frac{\mu_0 i}{2 \pi (a + x)}$.Consider a small element of length $dx$ at a distance $x$ from the pivot point of the rod.The velocity of this element is $v = x\omega$.The induced $emf$ $d\varepsilon$ across this small element is $d\varepsilon = B v dx = \left( \frac{\mu_0 i}{2 \pi (a + x)} \right) (x\omega) dx$.Integrating from $x = 0$ to $x = l$:$\varepsilon = \int_0^l \frac{\mu_0 i \omega}{2 \pi} \frac{x}{a + x} dx = \frac{\mu_0 i \omega}{2 \pi} \int_0^l \left( 1 - \frac{a}{a + x} \right) dx$.$\varepsilon = \frac{\mu_0 i \omega}{2 \pi} \left[ x - a \ln(a + x) \right]_0^l$.$\varepsilon = \frac{\mu_0 i \omega}{2 \pi} \left[ (l - a \ln(a + l)) - (0 - a \ln(a)) \right]$.$\varepsilon = \frac{\mu_0 i \omega}{2 \pi} \left[ l - a \ln \left( \frac{a + l}{a} \right) \right]$.

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