The magnetic field due to a straight conducto | vedclass

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Question

(A) For a straight cylindrical conductor of radius $a$ carrying a steady current $I$:$1$. Inside the conductor $(r < a)$: The magnetic field $B$ is di

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Vedclass · Answer

(A) For a straight cylindrical conductor of radius $a$ carrying a steady current $I$:$1$. Inside the conductor $(r $2$. Outside the conductor $(r \ge a)$: The magnetic field $B$ is inversely proportional to the distance $r$ from the axis,given by $B = \frac{\mu_0 I}{2 \pi r}$. This represents a hyperbolic decrease.Combining these,the graph shows a linear increase up to $r = a$ and a hyperbolic decrease for $r > a$. This matches the graph in option $A$.

The magnetic field due to a straight conductor of uniform cross section of radius $a$ and carrying a steady current is represented by

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