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

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:

& e=Blv \\\ & v=\omega l \\\ \end{aligned}$$ **Complete step-by-step solution:** From the given information, we have the data as follows. The potential difference between the end points P and Q is, 100 V. The expression for the emf induced in terms of the magnetic field, the length of the rod/material and the linear velocity of the electron is given as follows. $$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{aligned} & E=B\omega \left[ \dfrac{{{l}^{2}}}{2} \right]_{0}^{L} \\\ & \therefore E=B\omega \dfrac{{{L}^{2}}}{2} \\\ \end{aligned}$$ We know the value of the emf induced, so, substitute the same. $$\begin{aligned} & 100=B\omega \dfrac{{{L}^{2}}}{2} \\\ & \Rightarrow B\omega {{L}^{2}}=2\times 100 \\\ & \therefore B\omega {{L}^{2}}=200 \\\ \end{aligned}$$ Integrate the equation, $$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{aligned} & 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{aligned}$$ We know the value of the $$B\omega {{L}^{2}}$$, so, substitute the same. $$\begin{aligned} & E=\dfrac{3}{8}(200) \\\ & \therefore E=75\,V \\\ \end{aligned}$$ $$\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.