Question

The emf induced between M and Q if the potential between P and Q is 100 V. M is the midpoint of P and Q.

Answer 1

Hint: The point M between the points P and Q is a mid point. Thus, the emf induced between the points M and Q will be less than the emf induced between the points P and Q. Using the relation between the induced emf and length of the rod, we will compute the emf induced between the end point Q and the mid point M.

Formula used:
\[\begin{align}
  & e=Blv \\
 & v=\omega l \\
\end{align}\]

Complete step-by-step solution:
From the given information, we have the data as follows.
The potential difference between the end points P and Q is, 100 V.
The expression for the emf induced in terms of the magnetic field, the length of the rod/material and the linear velocity of the electron is given as follows.
\[e=Blv\]
The small change in the emf induced is given by considering the small length of material ‘dl’.
\[de=Bvdl\]
The relation between the linear velocity and the angular velocity is given as follows.
\[v=\omega l\]
Combine the above two equations. So, we get,
\[de=B(\omega l)dl\]
Integrate the above equation, taking the limitation for the small induced emf to be 0 to E and the limitation of the length of the conductor to be from 0 to L, as in this case, we are considering the whole conductor.
\[\int\limits_{0}^{E}{de}=\int\limits_{0}^{L}{B(\omega l)dl}\]
Continue further computation.
\[\begin{align}
  & E=B\omega \left[ \dfrac{{{l}^{2}}}{2} \right]_{0}^{L} \\
 & \therefore E=B\omega \dfrac{{{L}^{2}}}{2} \\
\end{align}\]
We know the value of the emf induced, so, substitute the same.
\[\begin{align}
  & 100=B\omega \dfrac{{{L}^{2}}}{2} \\
 & \Rightarrow B\omega {{L}^{2}}=2\times 100 \\
 & \therefore B\omega {{L}^{2}}=200 \\
\end{align}\]
Integrate the equation, \[de=B(\omega l)dl\]
taking the limitation for the small induced emf to be 0 to E and the limitation of the length of the conductor to be from 0 to \[\dfrac{L}{2}\], as in this case, we are considering the half of the conductor.
\[\int\limits_{0}^{E}{de}=\int\limits_{0}^{\dfrac{L}{2}}{B(\omega l)dl}\]
Continue further computation.
\[\begin{align}
  & E=B\omega \left[ \dfrac{{{l}^{2}}}{2} \right]_{0}^{\dfrac{L}{2}} \\
 & \Rightarrow E=B\omega \left[ \dfrac{{{L}^{2}}}{2}-\dfrac{{{L}^{2}}}{8} \right] \\
 & \Rightarrow E=B\omega \dfrac{3{{L}^{2}}}{8} \\
 & \therefore E=\dfrac{3}{8}B\omega {{L}^{2}} \\
\end{align}\]
We know the value of the \[B\omega {{L}^{2}}\], so, substitute the same.
\[\begin{align}
  & E=\dfrac{3}{8}(200) \\
 & \therefore E=75\,V \\
\end{align}\]
\[\therefore \] The emf induced between M and Q if the potential between P and Q is 100 V and M being the midpoint of P and Q is 75 V.

Note: The point M between the points P and Q is a mid point. Thus, the emf induced between the points M and Q will be less than the emf induced between the points P and Q, but not equal to exactly half the value.