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  • A homogeneous cylindrical rod of radius R carries a current I whose current density J (defined as current per unit cross-sectional area) is constant throughout the rod. The magnetic field at a

A homogeneous cylindrical rod of radius R carries a current I whose current density J (defined as current per unit cross-sectional area) is constant throughout the rod. The magnetic field at a point P at a distance r = R/2 from the centre of the rod is.

Option: 1

\mathrm{zero }


Option: 2

\mathrm{\frac{\mu_0 I}{\pi R}}


Option: 3

\mathrm{\frac{\mu_0 I}{2 \pi R}}


Option: 4

\mathrm{\frac{\mu_0 I}{4 \pi R}}


Answers (1)

best_answer

The current density \mathrm{J=\frac{I}{\pi R^2}}.
For a homogeneous cylinder, the current density J is constant and does not depend on the radial distance r of the point P from centre O of the cylinder (see Fig.)

Since the point P is inside the cylinder, r < R, the current enclosed by the Amperean loop is
i = J x Area of Amperean loop
\mathrm{ =\frac{I}{\pi R^2} \times \pi r^2=\frac{I r^2}{R^2} }

From Amperean Circuital law,
\mathrm{ \oint \mathbf{B} \cdot \mathbf{d} \mathbf{I} =\mu_0 i \\ }
or  \mathrm{ B \times 2 \pi r =\mu_0 \times \frac{I r^2}{R^2} \\ }
    \mathrm{ \Rightarrow \quad B =\frac{\mu_0 I r}{2 \pi R^2} }    .......(1)

Putting \mathrm{ r=\frac{R}{2} }     in eq. (1), we get
\mathrm{ B=\frac{\mu_{\mathrm{o}} I}{4 \pi R} }, which is choice (4).

Posted by

HARSH KANKARIA

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