A rectangular loop with a sliding connector of length=1.0m is situated in a uniform magnetic field field=2t perpendicular to the plane of loop. Resistance of the connector is r=2Ω . Two resistance is 6Ω and 3Ω are connected as shown in the figure. The external force required to keep the connector moving with a constant velocity v=2ms−1

In this question it is given that,

the resistance of the connector 2Ω

magnetic field – 2t

two resistance – 6Ω and 3Ω velocity of moving connector – 2ms−1

Let’s calculate the induced emf generated in the circuit. E = vBl

E = 2×2×1

E = 4V

Now, consider the two resistances 6Ω and 3Ω since they are in parallel connected then the resultant resistance will be 2Ω

We canfind the value of I (current) which is flowing through the circuit which is I = 4/2+2 which will give us the value 1ampere is flowing in clockwise direction. Now, calculation the magnetic force on connector by using formula

F = ilB

F = 1×1×2

⇒ F=2N The direction of this force is towards the left.

So, as a result, if we want to move our connector with a constant velocity, we need to apply force of 2N

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