Question

The wire PQ has mass 0.5 kg and resistance 2 Ω, can slide on the smooth, very long horizontal parallel rails separated by a distance of 1 m. The resistance of the rails is negligible. A uniform magnetic field of magnitude 0.6 T exists in the rectangular region, and a resistance of 3 Ω connects the rails outside the field region. At t=0, the wire PQ is pushed towards the right with a speed v=5 m/s. Find the maximum distance the wire will move.

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Solution

The correct option is **D** 34.7 m

Emf induced across the wire PQ is,

E=Bvl=0.6×v×1=0.6v

∴ Current flowing in the circuit is,

i=BvlR+r=0.6v5=0.12v

So, the force acting on the wire due to magnetic field is,

FB=ilB=0.12v×1×0.6=0.072v

∴ Acceleration of the wire is,

a=−FBm=−0.072v0.5=−0.144v

a is negative because velocity of wire is decreasing.

We know that,

a=vdvdx, where x is the displacement of the wire.

∴vdvdx=−0.144v

⇒dv=−0.144dx

Integrating both sides,

∫05dv=−∫x00.144dx

⇒[0−5]=−0.144x

∴x=34.7 m

Hence, (D) is the correct answer.

Emf induced across the wire PQ is,

E=Bvl=0.6×v×1=0.6v

∴ Current flowing in the circuit is,

i=BvlR+r=0.6v5=0.12v

So, the force acting on the wire due to magnetic field is,

FB=ilB=0.12v×1×0.6=0.072v

∴ Acceleration of the wire is,

a=−FBm=−0.072v0.5=−0.144v

a is negative because velocity of wire is decreasing.

We know that,

a=vdvdx, where x is the displacement of the wire.

∴vdvdx=−0.144v

⇒dv=−0.144dx

Integrating both sides,

∫05dv=−∫x00.144dx

⇒[0−5]=−0.144x

∴x=34.7 m

Hence, (D) is the correct answer.

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