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JEE Advanced 2012Paper 1
Exam
JEE Advanced
Year
2012
Paper
Paper 1
Subject
Physics
Chapter
Magnetic Effects of Current

JEE Advanced 2012 · Paper 1 · Physics · Magnetic Effects of Current

A cylindrical cavity of diameter a exists inside a cylinder of...

A cylindrical cavity of diameter a exists inside a cylinder of diameter 2a2 a as shown in the figure. Both the cylinder and the cavity are infinity long. A uniform current density JJ flows along the length. If the magnitude of the magnetic field at the point PP is given by N12μ0aJ\frac{N}{12} \mu_{0} a J, then the value of NN is A cylindrical cavity of diameter a exists inside a cylinder of diameter 2 a as shown in the figure. Both the cylinder

Correct answer

0

Step-by-step solution

Current density J= current  area =IA⇒I=JAJ=\frac{\text{ current }}{\text{ area }}=\frac{I}{A} \Rightarrow I=\mathrm{JA} Magnetic field BR\mathrm{B}_{\mathrm{R}} after removing cavity (C)(\mathrm{C}) BR=Btotal −Bcavity \mathrm{B}_{\mathrm{R}}=\mathrm{B}_{\text{total }}-\mathrm{B}_{\text{cavity }} μ0It2πa−μ0Ic2π(32a)=μ0πa[It2−Ic3]( here It=J(πa2)Ic=J(πa24))=μ0πa[πa2J2−πa2J12] or, BR=5μ0aJ12\begin{array}{l} \frac{\mu_{0} I_{t}}{2 \pi a}-\frac{\mu_{0} I_{c}}{2 \pi\left(\frac{3}{2} a\right)} \\ \quad=\frac{\mu_{0}}{\pi a}\left[\frac{I_{t}}{2}-\frac{I_{c}}{3}\right]\left(\text{ here } I_{t}=J\left(\pi a^{2}\right) I_{c}=J\left(\frac{\pi a^{2}}{4}\right)\right) \\ =\frac{\mu_{0}}{\pi a}\left[\frac{\pi a^{2} J}{2}-\frac{\pi a^{2} J}{12}\right] \\ \text{ or, } \mathrm{B}_{\mathrm{R}}=\frac{5 \mu_{0} a J}{12} \end{array} Comparing it with N12μ0aJ\frac{N}{12} \mu_{0} a J We get N=5N=5
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