Qstnid -Iv-49-figure Shows A Smooth Pair Of Thick Metallic Rails Connected Across A Battery Of Emf Ε Having A Negligible Internal Resistance. A Wire Ab Of Length L And Resistance R Can Slide Smoothly On The Rails. The Entire System Lies In A Horizontal Plane And Is Immersed In A Uniform Vertical Magnetic Field B. At An Instant T, The Wire Is Given A Small Velocity V Towards Right. (A) Find The Current In It At This Instant. What Is The Direction Of The Current? (B) What Is The Force Acting On The Wire At This Instant? (C) Show That After Some Time The Wire Ab Will Slide With A Constant Velocity. Find This Velocity. <Span Style="background-color: #Ffff00;">figure</span></span> - 38: Electromagnetic Induction - Neet-xii-physics

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NEET-XII-Physics

38: Electromagnetic Induction

with Solutions - page 10

Qstn# iv-49 Prvs-Qstn Next-Qstn

#49
Figure shows a smooth pair of thick metallic rails connected across a battery of emf ε having a negligible internal resistance. A wire ab of length l and resistance r can slide smoothly on the rails. The entire system lies in a horizontal plane and is immersed in a uniform vertical magnetic field B. At an instant t, the wire is given a small velocity v towards right. (a) Find the current in it at this instant. What is the direction of the current? (b) What is the force acting on the wire at this instant? (c) Show that after some time the wire ab will slide with a constant velocity. Find this velocity.
Figure
Ans : According to Fleming's left hand rule the force in the wire ab will be in the upward direction.
Moreover, a moving wire ab is equivalent to a battery of emf vBl as shown in the figure.

At the given instant, the net emf across the wire (d) is E - Bvl. (a) The current through the wire is given by
`` i=\frac{E-Bvl}{r}``
The direction of the current is from b to a. (b) The force acting on the wire at the given instant is given by
`` F=ilB=\left(\frac{E-Bvl}{r}\right)`` towards right (c) The velocity of the wire attains a value such that it satisfies E = Bvl.
The net force on the wire becomes zero. Thus, the wire moves with a constant velocity v.
∴ `` v=\frac{E}{Bl}``
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#49-a
Find the current in it at this instant. What is the direction of the current?
Ans : The current through the wire is given by
`` i=\frac{E-Bvl}{r}``
The direction of the current is from b to a.

#49-b
What is the force acting on the wire at this instant?
Ans : The force acting on the wire at the given instant is given by
`` F=ilB=\left(\frac{E-Bvl}{r}\right)`` towards right

#49-c
Show that after some time the wire ab will slide with a constant velocity. Find this velocity.
Figure
Ans : The velocity of the wire attains a value such that it satisfies E = Bvl.
The net force on the wire becomes zero. Thus, the wire moves with a constant velocity v.
∴ `` v=\frac{E}{Bl}``
Page No 310: