In the above diagram, a strong bar magnet is moving towards solenoid-2 from solenoid-1. The direction of induced current in solenoid-1 and that in solenoid-2, respectively, are through the directions :
When a strong bar magnet moves towards a solenoid, it induces an electromotive force (EMF) in the solenoid due to electromagnetic induction. The direction of the induced current can be determined using Lenz's Law, which states that the direction of the induced current is always such that it opposes the change in magnetic flux that produced it. Let's analyze the problem step by step:
Therefore, the correct option is AB and DC, where the induced current in solenoid-1 flows in direction AB and in solenoid-2 flows in direction DC respectively.
Step 1: Use Lenz’s Law
According to Lenz’s law, the induced current always flows in a direction such that it opposes the change in magnetic flux that causes it.
Step 2: Analyze Solenoid-1
As the bar magnet approaches solenoid-1, the magnetic flux increases. To oppose this increase, solenoid-1 generates a magnetic field opposite to the bar magnet’s field, inducing a current in the direction AB.
Step 3: Analyze Solenoid-2
For solenoid-2, the magnetic field due to the approaching bar magnet increases in the direction towards solenoid-1. To oppose this, solenoid-2 generates a magnetic field pointing away, inducing a current in the direction DC.
The current passing through the battery in the given circuit, is:
A bob of heavy mass \(m\) is suspended by a light string of length \(l\). The bob is given a horizontal velocity \(v_0\) as shown in figure. If the string gets slack at some point P making an angle \( \theta \) from the horizontal, the ratio of the speed \(v\) of the bob at point P to its initial speed \(v_0\) is :
A full wave rectifier circuit with diodes (\(D_1\)) and (\(D_2\)) is shown in the figure. If input supply voltage \(V_{in} = 220 \sin(100 \pi t)\) volt, then at \(t = 15\) msec: