Figure shows the cross-sectional view of the | vedclass

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(d) By using $B = \frac{{{\mu _0}i}}{{2\pi \,r}}\left( {\frac{{{r^2} - {a^2}}}{{{b^2} - {a^2}}}} \right)$ here $r = \frac{{3R}}{2},$$a = R,$$b = 2R$

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$\frac{{5{\mu _0}i}}{{36\pi \,R}}$

Vedclass · Answer

(d) By using $B = \frac{{{\mu _0}i}}{{2\pi \,r}}\left( {\frac{{{r^2} - {a^2}}}{{{b^2} - {a^2}}}} ight)$ here $r = \frac{{3R}}{2},$$a = R,$$b = 2R$
$ \Rightarrow $ $B = \frac{{{\mu _0}i}}{{2\pi \left( {\frac{{3R}}{2}} ight)}} 	imes \left\{ {\frac{{{{\left( {\frac{{3R}}{2}} ight)}^2} - {R^2}}}{{(2{R^2}) - {R^2}}}} ight\}$$ = \frac{{5.{\mu _o}i}}{{36\pi \,r}}$.

Figure shows the cross-sectional view of the hollow cylindrical conductor with inner radius '$R$' and outer radius '$2R$', cylinder carrying uniformly distributed current along it's axis. The magnetic induction at point '$P$' at a distance $\frac{{3R}}{2}$ from the axis of the cylinder will be

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