Question
As shown in the figure, $$P$$ and $$Q$$ are two coaxial conducting loops separated by some distance. When the switch $$S$$ is closed, a clockwise current $${I_P}$$ flows in $$P$$ (as seen by $$E$$) and an induced current $${I_{{Q_1}}}$$ flows in $$Q.$$ The switch remains closed for a long time. When $$S$$ is opened, a current $${I_{{Q_2}}}$$ flows in $$Q.$$ Then the direction $${I_{{Q_1}}}$$ and $${I_{{Q_2}}}$$ (as seen by $$E$$) are
As shown in the figure, $$P$$ and $$Q$$ are two coaxial conducting loops separated by some distance. When the switch $$S$$ is closed, a clockwise current $${I_P}$$ flows in $$P$$ (as seen by $$E$$) and an induced current $${I_{{Q_1}}}$$ flows in $$Q.$$ The switch remains closed for a long time. When $$S$$ is opened, a current $${I_{{Q_2}}}$$ flows in $$Q.$$ Then the direction $${I_{{Q_1}}}$$ and $${I_{{Q_2}}}$$ (as seen by $$E$$) are
A.
respectively clockwise and anti-clockwise
B.
both clockwise
C.
both anti-clockwise
D.
respectively anti-clockwise and clockwise
Answer :
respectively anti-clockwise and clockwise
Solution :
When switch $$S$$ is closed, a magnetic field is set-up in the space around $$P.$$ The field lines threading $$Q$$ increases in the direction from right to left. According to Lenz's law, $${I_{{Q_1}}}$$ will flow so as to oppose the cause and flow in anticlockwise direction as seen by $$E.$$ Reverse is the case when $$S$$ is opened. $${I_{{Q_2}}}$$ will be clockwise.
When switch $$S$$ is closed, a magnetic field is set-up in the space around $$P.$$ The field lines threading $$Q$$ increases in the direction from right to left. According to Lenz's law, $${I_{{Q_1}}}$$ will flow so as to oppose the cause and flow in anticlockwise direction as seen by $$E.$$ Reverse is the case when $$S$$ is opened. $${I_{{Q_2}}}$$ will be clockwise.
