IIT-JEE 2012 · Paper 1 · Physics Q17

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PhysicsInteger+4Question 17 of 60

A cylindrical cavity of diameter $a$ exists inside a cylinder of diameter $2a$ 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 $\dfrac N{12}\mu_0aJ$ , then the value of $N$ is

Cross-section: a shaded circle of diameter 2a centred at O with a circular cavity of diameter a touching the right edge (cavity centre a/2 to the right of O); point P is on the left edge of the big circle, on the line through O and the cavity centre.

Your answer

Answer and worked solution

Official answer: 5

Superposition. Full cylinder (radius $a$ ) with current density $J$ , plus the cavity (radius $a/2$ ) carrying $-J$ .

Full cylinder at P (on its surface, distance $a$ ): $B_1=\dfrac{\mu_0J\pi a^2}{2\pi a}=\dfrac{\mu_0Ja}2$ .

Cavity. Its centre is $a/2$ from O on the far side, so P is $\frac{3a}2$ from it:
$B_2=\frac{\mu_0J\pi(a/2)^2}{2\pi(3a/2)}=\frac{\mu_0Ja}{12}.$
Both fields at P are along the same line and opposite.

Net. $B=\mu_0Ja\left(\frac12-\frac1{12}\right)=\frac5{12}\mu_0aJ$ , so $N=5$ .

Answer: 5

Topics:Ampere's circuital law

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IIT-JEE 2012 · Paper 1, Physics question 17