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Q.

A cylindrical conductor of radius $R$ carries a current along its length. The current density $J$ varies according to distance $x$ from axis given by, $J = J_{0} (1 - \frac{x}{R})$ where $J_{0} π$ is a constant. If value of maximum magnetic field intensity is $B_{0}$ at $x = d_{0}$ , then

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a

$d_{0} = \frac{3 R}{4}$

b

$d_{0} = \frac{3 R}{8}$

c

$B_{0} = \frac{3 μ_{0} J_{0} R}{16}$

d

Total current through conductor is $\frac{π J_{0} R^{2}}{3}$

answer is A, C, D.

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Detailed Solution

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Apply Ampere’s circuital law

$\begin{array}{l} I = \int_{0}^{R} J (x) 2 π x d x \\ = \frac{π J_{0} R^{2}}{3} \\ B .2 π x = μ_{0} J_{0} \int_{0}^{x} (1 - \frac{x}{R}) 2 π x d x \\ B = μ_{0} J_{0} (\frac{x}{2} - \frac{x^{2}}{3 R}) \\ \frac{d B}{d x} = 0 \Rightarrow x = \frac{3 R}{4} \end{array}$

Maximum value of magnetic field is $B_{0} = \frac{3 μ_{0} J_{0} R}{16}$

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