A very long straight conductor and an isoscel | vedclass

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(D) Consider an element of width $dx$ of the triangular conductor at a distance $x$ from its vertex. The height of the triangle is $h$ and the base is

Vedclass · Accepted Answer

$1.2 	imes 10^{-8} \ H$

Vedclass · Answer

(D) Consider an element of width $dx$ of the triangular conductor at a distance $x$ from its vertex. The height of the triangle is $h$ and the base is $b$. By similar triangles,the width of the strip at distance $x$ is $w = (b/h)x$.The area of this strip is $dA = (b/h)x \, dx$.The magnetic field $B$ due to the long straight wire at a distance $(a+x)$ is $B = \frac{\mu_0 I}{2\pi(a+x)}$.The magnetic flux $d\phi$ through the strip is $d\phi = B \, dA = \frac{\mu_0 I}{2\pi(a+x)} \cdot \frac{b}{h} x \, dx$.Integrating from $x = 0$ to $x = h$:$\phi = \int_0^h \frac{\mu_0 I b}{2\pi h} \frac{x}{a+x} \, dx = \frac{\mu_0 I b}{2\pi h} \int_0^h \left( 1 - \frac{a}{a+x} ight) \, dx$.$\phi = \frac{\mu_0 I b}{2\pi h} [x - a \ln(a+x)]_0^h = \frac{\mu_0 I b}{2\pi h} [h - a \ln((a+h)/a)]$.The coefficient of mutual induction $M = \phi/I = \frac{\mu_0 b}{2\pi h} [h - a \ln((a+h)/a)]$.Substituting values: $\mu_0 = 4\pi 	imes 10^{-7} \ T \cdot m/A$,$a = 0.1 \ m$,$b = 0.2 \ m$,$h = 0.1 \ m$.$M = \frac{4\pi 	imes 10^{-7} 	imes 0.2}{2\pi 	imes 0.1} [0.1 - 0.1 \ln(0.2/0.1)] = 2 	imes 2 	imes 10^{-7} [0.1 - 0.1 \ln(2)] = 4 	imes 10^{-7} 	imes 0.1 [1 - 0.693] = 4 	imes 10^{-8} 	imes 0.307 \approx 1.228 	imes 10^{-8} \ H$.

$A$ very long straight conductor and an isosceles triangular conductor lie in a plane and are separated from each other as shown in the figure. If $a = 10 \ cm$,$b = 20 \ cm$,and $h = 10 \ cm$,find the coefficient of mutual induction.

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