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Question: When a 'J' shaped conducting rod is rotating in its own plane with constant angular velocity $\omega...

When a 'J' shaped conducting rod is rotating in its own plane with constant angular velocity ω\omega, about one of its ends P, in a uniform magnetic field B as shown then the magnitude of induced emf across it will be

A

BωL2+l2B \omega \sqrt{L^2 + l^2}

B

12BωL2\frac{1}{2}B \omega L^2

C

12Bω(L2+l2)\frac{1}{2}B \omega (L^2 + l^2)

D

12Bωl2\frac{1}{2}B \omega l^2

Answer

(C)

Explanation

Solution

The induced electromotive force (EMF) across a conductor rotating in a uniform magnetic field is given by the formula:

E=12Bω(r22−r12)\mathcal{E} = \frac{1}{2} B \omega (r_2^2 - r_1^2)

where:

  • BB is the magnitude of the uniform magnetic field.
  • ω\omega is the constant angular velocity.
  • r1r_1 is the distance of one end of the conductor from the pivot point.
  • r2r_2 is the distance of the other end of the conductor from the pivot point.

In this problem, the 'J' shaped conducting rod is rotating about its end P. Therefore, P is the pivot point. So, for point P, r1=0r_1 = 0.

Let Q be the other end of the J-shaped rod. We need to find the straight-line distance from P to Q, which will be r2r_2. From the diagram:

  1. There is a straight segment of the rod starting from P, with length ll. Let's call the end of this straight segment R. So, the distance PR = ll.
  2. The curved part of the rod starts from R and ends at Q. The dimension 'L' is shown as the straight-line distance from R to Q, perpendicular to the segment PR.

Therefore, P, R, and Q form a right-angled triangle with the right angle at R. The sides of this right-angled triangle are PR = ll and RQ = LL. The straight-line distance from P to Q, which is the hypotenuse of this triangle, is r2=PQr_2 = \text{PQ}. Using the Pythagorean theorem:

r2=(PR)2+(RQ)2=l2+L2r_2 = \sqrt{(PR)^2 + (RQ)^2} = \sqrt{l^2 + L^2}

Now, substitute r1=0r_1 = 0 and r2=l2+L2r_2 = \sqrt{l^2 + L^2} into the EMF formula:

E=12Bω((l2+L2)2−02)\mathcal{E} = \frac{1}{2} B \omega ((\sqrt{l^2 + L^2})^2 - 0^2) E=12Bω(l2+L2)\mathcal{E} = \frac{1}{2} B \omega (l^2 + L^2)

This formula indicates that the induced EMF in a rotating conductor depends only on the straight-line distance of its ends from the center of rotation, not on the actual path or shape of the conductor.