JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Magnetic Effects of Current

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

Answer: 5

Solution

Using superposition, the field at

P

is the vector sum of the field due to the solid cylinder and the negative field due to the cavity. For the solid cylinder at distance

a

from its axis:

B_{1} = \frac{μ _{0} J π a ^{2}}{2 π a} = \frac{μ _{0} J a}{2}

. For the cavity (radius

a /2

) at distance

3 a /2

from

P

B_{2} = \frac{μ _{0} J π ( a /2 ) ^{2}}{2 π ( 3 a /2 )} = \frac{μ _{0} J a}{12}

. The net field is

B = B_{1} - B_{2} = \frac{μ _{0} J a}{2} - \frac{μ _{0} J a}{12} = \frac{5 μ _{0} J a}{12}

. Comparing with

\frac{N μ _{0} a J}{12}

, we get

N = 5

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Integer answer