Figures A and B show two long straight wires | vedclass

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(C) For a long straight wire of radius $R$ carrying a uniformly distributed current $I$,the magnetic field $B$ at a distance $r$ from the axis is give

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(C) For a long straight wire of radius $R$ carrying a uniformly distributed current $I$,the magnetic field $B$ at a distance $r$ from the axis is given by:Inside the wire $(r Outside the wire $(r \ge R)$: $B = \frac{\mu_0 I}{2 \pi r}$,which implies $B \propto \frac{1}{r}$.The maximum magnetic field occurs at the surface $(r = R)$ and its value is $B_{max} = \frac{\mu_0 I}{2 \pi R}$.Since $a Because $a  B_{max, b}$.Thus,the graph for wire $a$ must reach a higher peak value at a smaller radius $r = a$,and the graph for wire $b$ must reach a lower peak value at a larger radius $r = b$. This corresponds to the graph where curve $a$ peaks higher and earlier than curve $b$.

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