1.0 m rectangular loop with a sliding connector is in uniform magnetic field 2t perpendicular to plane of loop. Resistance is 2 ohms. Determine force.

Now, as per the question, the external force needed to keep connector moving with constant velocity equals 2ms−1. With the given data, we have to determine force.

Accoding to the question, the following can be said:

Resistance of Connector = 2Ω
Given Magnetic Field = 2t
Other Two Resistances = 6Ω and 3Ω (velocity of moving connector is 2ms⁻¹)

Now, in order to estimate the induced emf generated in the circuit, E = vBl

⇒ E = 2 × 2 × 1

⇒ E = 4V

Thus, after assuming the two resistances 6Ω and 3Ω because they’re parallelly connected, then the resultant resistance should be 2Ω.

Therefore, the value of I (current) that flows via the circuit is, I = 4/2 + 2, giving the value of 1 Ampere (flowing in a clockwise direction). Now, in order to estimate the magnetic force on the connector by using the formula:

F = ilB

After replacing the values, we get,

F = 1 × 1 × 2

⇒ F = 2N, (Thus, the direction of force is towards left).

Thus, it can be said that if we want to move our connector with a constant velocity, a force of 2N is needed to be applied.

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1.0 m rectangular loop with a sliding connector is in uniform magnetic field 2t perpendicular to plane of loop. Resistance is 2 ohms. Two resistances, 6 ohms and 3 ohms, are connected.

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1.0 m rectangular loop with a sliding connector is in uniform magnetic field 2t perpendicular to plane of loop. Resistance is 2 ohms. Two resistances, 6 ohms and 3 ohms, are connected.

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