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Question

Obtain an expression for the magnetic induction at a point due to an infinitely long straight conductor carrying current.


Solution

Magnetic induction due to infinitely long straight conductor carrying current: XY, is an infinitely long straight conductor carrying a current $$I$$(Figure).
$$P$$ is a point at a distance $$a$$ from the conductor. $$AB$$ is a small element of length $$dl$$. $$\theta$$ is the angle between the current element $$I dl$$ and the line joining the element $$dl$$ and the point P. According to Biot-Savart law, the magnetic induction at the point P due to the current element $$Idl$$ is.
$$dB=\displaystyle\dfrac{\mu_o}{4\pi}\dfrac{Idl\cdot\sin\theta}{r^2}$$          ......$$(1)$$
AC is drawn perpendicular to BP from A.
$$\angle OPA=\phi$$, $$\angle APB=d\phi$$
In $$\Delta$$ABC, $$\sin\theta=\dfrac{AC}{AB}=\dfrac{AC}{dl}$$
$$\therefore AC=dl\sin\theta$$           ..........$$(2)$$
From $$\Delta$$APC, AC$$=rd\phi$$         .........$$(3)$$
From equations $$(2)$$ and $$(3)$$,
$$rd\phi =dl\sin\theta$$         ............$$(4)$$
Substituting equation $$(4)$$ in equation $$(1)$$
$$dB=\displaystyle\dfrac{\mu_o}{4\pi}\dfrac{Ird\phi}{r^2}=\dfrac{\mu_o}{4\pi}\dfrac{Id\phi}{r}$$              .............$$(5)$$
In $$\Delta$$OPA, $$\cos\phi=\dfrac{a}{r}$$
$$\therefore r=\displaystyle\dfrac{a}{cos\phi}$$            ..........$$(6)$$
Substituting equation $$(6)$$ in equation $$(5)$$
$$dB=\displaystyle\dfrac{\mu_o}{4\pi}\dfrac{1}{a}\cos\phi d\phi$$
The total magnetic induction at P due to the conductor XY is
$$B=\overset{\phi_2}{\underset{-\phi_1}{\int}}dB=\overset{\phi_2}{\underset{-\phi_1}{\int}}\dfrac{\mu_o I}{4\pi a}\cos \phi d\phi$$
$$B=\dfrac{\mu_oI}{4\pi a}[\sin\phi_1+\sin\phi_2]$$
For infinitely long conductor,
$$\phi_1=\phi_2=90^o$$
$$\therefore B=\dfrac{\mu_oI}{2\pi a}$$
If the conductor is places in a medium of permeability $$\mu_0$$, $$B=\dfrac{\mu I}{2\pi a}$$.

633696_606186_ans_f09d70726e2e4db9a3124963acab8bb4.png

Physics

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