First slide

Ampere's circuital law

Question

A long straight metal rod has a very long hole of radius drilled parallel to the rod axis as shown in the figure. If the rod carries a current I, find the magnetic field on axis of hole. Given C is the centre of the hole and OC = c.

Difficult

A

$\frac{μ_{0} ic}{π (b^{2} - a^{2})}$
B

$\frac{μ_{0} ic}{2 π (b^{2} - a^{2})}$
C

$\frac{μ_{0} i (b^{2} - a^{2})}{2 πc}$
D

$\frac{μ_{0} ic}{2 π b^{2}}$

Solution

In the rod, current density
$j = \frac{i}{π (b^{2} - a^{2})}$
Actual field at any point in the cylindrical cavity is the vector sum of two current carrying rods, having same current density but in opposite direction. On the hole axis, only the larger rod contributes magnetic field.
$\begin{array}{l} \oint \vec{B} . \vec{dl} = μ_{0} i_{enclosed} \\ B 2 πc = μ_{0} ({jπc}^{2}) \\ B = \frac{μ_{0} {iπc}^{2}}{2 πcx (b^{2} - a^{2})} = \frac{μ_{0} ic}{2 π (b^{2} - a^{2})} \\ B = \frac{μ_{0} ic}{2 π (b^{2} - a^{2})} \end{array}$

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