# Electromagnetic induction with light bulbs

I chose C as the voltage, which was initially split between the two lamps, will only be "used" by lamp 1, thus increasing the brightness. However, the answer is D.

• On first reading, I would have chosen C as the correct answer too. Oct 18, 2016 at 1:12
• Do you know why the answer is D? Oct 18, 2016 at 1:59
• I think that the answer is C, too.
– user93237
Oct 18, 2016 at 6:50

Assume that the copper wire a very much smaller resistance than that if the bulbs and that the resistance of the bulbs $R$ is independent of the current passing through them.
Let the induced emf in the loop be $2\mathcal E$.
The current in the circuit which produces a certain equal brightness for the two bulbs is $I = \frac{2\mathcal E}{2R}=\frac{\mathcal E}{R}$.

Suppose the whole area of the loop was $4A$.
With the connecting wire as shown in the right hand diagram let the area of the top loop be $3A$ and that of the bottom loop be $A$.
The emf induced in the top loop is now $1.5\mathcal E$ because the rate of change of magnetic flux in the top loop has increased. And the current $\frac{1.5\mathcal E}{R}$ which is larger than before so the top bulb is brighter.
For the bottom loop the current is $\frac{0.5\mathcal E}{R}$ which is smaller than before so the bottom bulb is less bright.
An alternative (quicker?) method. Now put the wire exactly down the middle so that the induced emf in each loop, top and bottom, is $\mathcal E$.
So the current through each loop is exactly the same as before $I =\frac{\mathcal E}{R}$ and so the bulbs are as bright as before.
The emf induced in the top loop is slightly less than $2\mathcal E$ and so the current through the top bulb is just less than $\frac{2\mathcal E}{R}$ which is slightly less than twice the current in the original circuit and so the top bulb is brighter than before.