A cylindrical cavity of diameter exists inside a cylinder of diameter as shown in the figure. Both the cylinder and the cavity are infinitely long. A uniform current density flows along the length. If the magnitude of the magnetic field at the point i

Question

A cylindrical cavity of diameter $a$ exists inside a cylinder of diameter $2 a$ as shown in the figure. Both the cylinder and the cavity are infinitely long. A uniform current density $J$ flows along the length. If the magnitude of the magnetic field at the point $P$ is given by $\frac{N}{12} μ_{0} a J,$ then the value of $N$ is [2012]

Smart Achievers · Accepted Answer

(5)

$Current density J = \frac{current}{area} = \frac{I}{A} \Rightarrow I = J A$

$Magnetic field B_{R} after removing cavity (C)$

$B_{R} = B_{total} - B_{cavity}$

$B_{R} = \frac{μ_{0} I_{t}}{2 π a} - \frac{μ_{0} I_{c}}{2 π (\frac{3}{2} a)}$

$= \frac{μ_{0}}{π a} [\frac{I_{t}}{2} - \frac{I_{c}}{3}]$ $(here I_{t} = J (π a^{2}), I_{c} = J (\frac{π a^{2}}{4}))$

$= \frac{μ_{0}}{π a} [\frac{π a^{2} J}{2} - \frac{π a^{2} J}{12}]$ or, $B_{R} = \frac{5 μ_{0} a J}{12}$

$Comparing it with \frac{N}{12} μ_{0} a J, we get N = 5$

Solution

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Solution

Ans. (5) Current density J=currentarea=IA⇒I=JA Magnetic field BR after removing cavity (C) BR=Btotal-Bcavity BR=μ0It2πa-μ0Ic2π(32a) =μ0πa[It2-Ic3] (here It=J(πa2), Ic=J(πa24)) =μ0πa[πa2J2-πa2J12] or, BR=5μ0aJ12 Comparing it with N12μ0aJ, we get N=5

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