A solid cylinder of radius R and having current density $\frac{i}{π R^{2}}$ has a cavity of radius $\frac{R}{2}$ as shown in figure (Top view). If magnetic field at point P which is at a distance of 2R from the axis of the cylinder towards cavity is $\frac{μ_{0} i}{X π r},$ fill value of x.

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answer is 6.

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Detailed Solution

$B P = B_{c y l i n d e r} - B_{c a v i t y} = \frac{μ_{0} \hat{i}}{2 π (2 R)} - \frac{μ \hat{i}}{2 π (\frac{3 R}{2})} = \frac{μ_{0}}{4 π R} [i] - \frac{μ_{0} i}{12 π R} = \frac{μ_{0} i (3 - 1)}{12 π R} = \frac{μ_{0} i}{6 π R}$

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