An infinitely long current carrying thick cylindrical conductor, has a cylindrical cavity of radius $\frac{R}{2}$ . The axis of the cylindrical cavity is at a distance $\frac{R}{2}$ from the axis of the cylindrical conductor. A current with current density J flows along the length of the conductor. The magnetic field at a point ‘P’ at a distance 2R from the axis of the cylinder is (point ‘P’ is on the plane containing both the axes)

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magnetic field at P is

\frac{μ_{0} JR}{6}

magnetic field on the axis of the cavity is non-zero

magnetic field at P is

\frac{μ_{0} JR}{12}

magnetic field on the axis of the conductor is non-zero

answer is B, C, D.

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Detailed Solution

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At (P) $B_{1}$ due to cylinder is $B_{1} = \frac{μ_{0} i_{1}}{2 π (2 R)} \otimes$

$B_{2} = \frac{μ_{0} i_{2}}{2 π (\frac{3 R}{2})} ⊙$

$i_{1} = J π R^{2} and i_{2} = J . \frac{π R^{2}}{4}$

$B_{P} = \frac{μ_{0} JR}{6}$

B on the axis of cavity due to cylinder is zero

B on the axis of cavity due to cylinder is $\frac{μ_{0} JR}{2}$ . Total field on the cavity is non – zero

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