Question

A conducting rod AC of length $4l$ and mass m is rotating about a point O with angular momentum J. A uniform magnetic field B directed into the paper. Given $AO = l$ and $OC = 3l$ . Then ${V_A} - {V_C}$ is

(A) $\dfrac{5}{{12}}\dfrac{{BJ}}{m}$
(B) $\dfrac{7}{{12}}\dfrac{{BJ}}{{m{l^2}}}$
(C) $\dfrac{{12}}{7}\dfrac{{BJ}}{m}$
(D) $\dfrac{{12}}{7}\dfrac{{BJ}}{{m{l^2}}}$

Answer 1

It is given that the rod is said to be rotating about a point with a specific angular momentum. In presence of magnetic field B, the rod rotates anti-clockwise with momentum. Now, we know that emf of any rotating rod is a product of magnetic field , angular velocity and its length. Using this, find the potential difference between the ends.

Complete Step by Step Solution
The rod is said to be under the influence of the constant magnetic field of intensity B. This causes the rod to rotate along the anti-clockwise direction with an angular velocity of magnitude $\omega$ . Now, the emf induced on the rod is given as ,
$\Rightarrow e = Blv$
Where, B is the magnetic field intensity, l is the length of the rod and v is the speed relative to the magnetic field.
Now, speed of rod can be written as,
$\Rightarrow \omega = \dfrac{v}{l}$
where $\omega$ is the angular velocity of the rod.
We can write angular velocity in terms of the angular momentum of the rod, since angular momentum of the rod is given as J.
$\Rightarrow J = I\omega$
Moment of inertia about point O is given as the sum of moment of inertia about OA and moment of inertia about OC
$\Rightarrow I = {I_{OA}} + {I_{OC}}$
$\Rightarrow I = \dfrac{1}{3}[\dfrac{{M{l^2}}}{4} + \dfrac{{3M}}{4}{(3l)^2}]$ (Since, moment of inertia of the rod is $\dfrac{1}{3}M{l^2}$ ]
$\Rightarrow I = \dfrac{1}{3}[\dfrac{{28M{l^2}}}{4}] = \dfrac{{7M{l^2}}}{3}$
Substituting this in angular momentum equation we get,
$\Rightarrow J = I\omega$
On, Rearranging we get,
$\Rightarrow \dfrac{J}{I} = \omega$
On substituting I,
$\Rightarrow \dfrac{{3J}}{{7M{l^2}}} = \omega$
Now, let us consider AO. The length is l and total length is 4l , we can calculated the emf experienced by the rod as ,
$\Rightarrow e = Blv = \dfrac{1}{2}B{l^2}\omega$
Now, for AO, we get,
$\Rightarrow {e_{OA}} = {e_o} - {e_A} = \dfrac{1}{2}B{l^2}\omega$
Considering length OC. The length is 3l and the total length of the rod is 4l. Emf dispersed by the rod is given as
$\Rightarrow {e_{OC}} = {e_o} - {e_C} = \dfrac{9}{2}B{l^2}\omega$
Now, emf difference is
$\Rightarrow {e_A} - {e_C} = 4B{l^2}\omega$
Substituting for $\omega$ , we get,
$\Rightarrow {e_A} - {e_C} = 4B{l^2} \times \dfrac{{3J}}{{7M{l^2}}}$
Cancelling the common term, we get,
$\Rightarrow {e_A} - {e_C} = \dfrac{{12BJ}}{{7M}}$
Hence, option (C) is the right answer.

Note:
The moment of inertia is also the rotational equivalence of mass. Now, for a road rotating about one end , the moment of inertia about the central portion or the concentrated portion is given as 1/3rd of the product of its mass and square of the length .