Experts explain the answer of this question
Dear Student,
If a current in loop P in a clockwise direction as per E, then according to Fleming's right hand thumb rule, the magnetic flux lines will go into the loop as per E.
Current will be induced in the loop Q only when the magnetic flux changes which will happen only when the current appears or disappears (as the current is dc). So, as per E, when the switch is closed there will be an increase in flux into the loop P and consequently into the loop Q also.
According to Lenz's law, a current will be induced in loop Q in such a direction that magnetic lux comes out of the loop Q (as per E) to compensate the increase. For E, an anticlockwise current IQ1 must set up in loop Q.
After some time this induced current becomes 0 as the change in magnetic flux due to the current in loop P becomes constant.
When the switch S is opened after a long time, the current in loop P decreases to 0. So, the magnetic flux due to this current into the loop P decreases with time. In order to compensate this decrease in magnetic flux, a current must be induced in loop Q which will try to increase the magnetic flux into the loop P and hence Q. In order to get a flux into the loop Q (as per E), the current induced in loop Q (IQ2) must be clockwise.
Thus, the direction of IQ1 and IQ2 must be anticlockwise and clockwise, respectively.
Ans (d)
Regards
If a current in loop P in a clockwise direction as per E, then according to Fleming's right hand thumb rule, the magnetic flux lines will go into the loop as per E.
Current will be induced in the loop Q only when the magnetic flux changes which will happen only when the current appears or disappears (as the current is dc). So, as per E, when the switch is closed there will be an increase in flux into the loop P and consequently into the loop Q also.
According to Lenz's law, a current will be induced in loop Q in such a direction that magnetic lux comes out of the loop Q (as per E) to compensate the increase. For E, an anticlockwise current IQ1 must set up in loop Q.
After some time this induced current becomes 0 as the change in magnetic flux due to the current in loop P becomes constant.
When the switch S is opened after a long time, the current in loop P decreases to 0. So, the magnetic flux due to this current into the loop P decreases with time. In order to compensate this decrease in magnetic flux, a current must be induced in loop Q which will try to increase the magnetic flux into the loop P and hence Q. In order to get a flux into the loop Q (as per E), the current induced in loop Q (IQ2) must be clockwise.
Thus, the direction of IQ1 and IQ2 must be anticlockwise and clockwise, respectively.
Ans (d)
Regards