Question

Two conducting wires are joined together and are moving in a magnetic field perpendicular to the plane of the wires, as shown in the figure. If $V_P$ and $V_Q$ are the potentials at points $P$ and $Q$ respectively, then -

• A. $V_P>V_Q$
• B. $V_P=V_Q$
• C. $V_P<V_Q$
• D. $V_P\le V_Q$

Answer 1

The correct option is A $V_P>V_Q$


From the figure, we can see that, $V_O>V_P$ using the right-hand rule and,

$V_{OP}=\frac{B\omega l^2}{2}=\frac{0.8\times 20\times 1^2}{2}=8\text{V}$

Similarly,

$V_{OQ}=\frac{B\omega l^2}{2}=\frac{0.8\times 20\times (1^2+2^2)}{2}=40\text{V}$

Now,

$V_{PQ}=V_P-V_Q=V_P-V_O-V_Q+V_O$

$\Rightarrow V_{PQ}=-(V_O-V_P)+(V_O-V_Q)=V_{OQ}-V_{OP}$

$\Rightarrow V_{PQ}=40-8=32\text{V}>0\Rightarrow V_P>V_Q$