Question
There are three wire $$MO,NO$$ and $$PQ,$$ wires $$MO$$ and $$NO$$ are fixed and perpendicular to each other. Wire $$PQ$$ moves with a constant velocity $$v$$ as shown in the figure and resistance per unit length of each wire is $$\lambda $$ and magnetic field exists perpendicular and inside the paper then Which of the following is wrong?
There are three wire $$MO,NO$$ and $$PQ,$$ wires $$MO$$ and $$NO$$ are fixed and perpendicular to each other. Wire $$PQ$$ moves with a constant velocity $$v$$ as shown in the figure and resistance per unit length of each wire is $$\lambda $$ and magnetic field exists perpendicular and inside the paper then Which of the following is wrong?
A.
current in loop is anticlockwise
B.
magnitude of current in the loop is $$\frac{{Bv}}{{\lambda \left( {\sqrt 2 + 1} \right)}}$$
C.
current in the loop is independent of time.
D.
magnitude of current decreases as time increases.
Answer :
magnitude of current decreases as time increases.
Solution :
$$\eqalign{ & \phi = BA = \frac{B}{2}\frac{x}{{\sqrt 2 }}\frac{x}{{\sqrt 2 }} \cr & \varepsilon = \frac{{d\phi }}{{dt}} = \frac{{2x}}{4}\frac{{dx}}{{dt}}B = \frac{x}{2}Bv \cr & \varepsilon = \frac{{d\phi }}{{dt}} = \frac{{2x}}{4}\frac{{dx}}{{dt}}B = \frac{x}{2}B\frac{{d\left( {2y} \right)}}{{dt}} \cr & i = \frac{{xBv}}{{2\lambda \left( {x + \sqrt {2x} } \right)}} = \frac{{Bv}}{{\lambda \left( {1 + \sqrt 2 } \right)}} \cr} $$
$$\eqalign{ & \phi = BA = \frac{B}{2}\frac{x}{{\sqrt 2 }}\frac{x}{{\sqrt 2 }} \cr & \varepsilon = \frac{{d\phi }}{{dt}} = \frac{{2x}}{4}\frac{{dx}}{{dt}}B = \frac{x}{2}Bv \cr & \varepsilon = \frac{{d\phi }}{{dt}} = \frac{{2x}}{4}\frac{{dx}}{{dt}}B = \frac{x}{2}B\frac{{d\left( {2y} \right)}}{{dt}} \cr & i = \frac{{xBv}}{{2\lambda \left( {x + \sqrt {2x} } \right)}} = \frac{{Bv}}{{\lambda \left( {1 + \sqrt 2 } \right)}} \cr} $$
