Figure (38-E23) 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 u 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.

Open in App

Solution

Net emf =E -Bvl

$i = \frac{E - B v l}{r}$ (from b to a)

$F = i l B = (\frac{E - B v l}{r})$

$l B = (\frac{l B}{r})$ (E -Bvl)towards right After some time,

when E =Bvl, then the wire moves in constant velocity v.

Hence $v = \frac{E}{B l}$

Suggest Corrections

Similar questions

Q. The figure shows a smooth pair of thick metallic rails connected across a battery of emf

ε

having a negligible internal resistance. A wire

a b

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 any instant, the wire is given a small velocity

v

towards the right. Find the current in the arrangement at this instant.

A conducting wire ab of length l, resistance r and mass m starts sliding at t = 0 down a smooth, vertical, thick pair of connected rails as shown in figure (38-E24). A

uniform magnetic field B exists in the space in a direction perpendicular to the plane of the rails (a) Write the induced emf in the loop at an instant ; when the speed of the wire is v. (b) What would be the magnitude and direction of the induced current in the wire ? (c) Find the downward acceleration of the wire at this instant. (d) After sufficient time, the wire starts moving with a constant velocity. Find this velocity nit (e) Find the velocity of the wire as a function of time ft) Find the displacement of the wire as a function ai time. (g) Show that the rate of heat developed in the wire is equal to the rate at which the gravitational potential energy is decreased after steady state is reached.

Q. Consider the situation shown in the figure. The wire PQ has mass m, resistance r and can slide on the smooth horizontal parallel rails of separated by a distance

l

. The resistance of the rails is negligible . A uniform magnetic field B exist in the rectangular region and a resistance R connects the rails outside the field region. At t=0, the wire PQ is pushed towards right with a speed of

v_{0}

. Find
(a) the current in the loop in an instant when the speed of the wire PQ is v.
(b) the acceleration of the wire at this instant
(c) Velocity v as a function

Q. Consider the situation shown in figure. The wire PQ has mass m, resistance r and can slide on the smooth, horizontal parallel rails separated by a distance l. The resistance of the rails is negligible. A uniform magnetic field B exists in the rectangular region and a resistance R connects the rails outside the field region. At t = 0, the wire PQ is pushed towards right with a speed v₀. Find (a) the current in the loop at an instant when the speed of the wire PQ is v, (b) the acceleration of the wire at this instant, (c) the velocity v as a functions of x and (d) the maximum distance the wire will move.

The current generator $I_{g}$ , shown in figure (38-E20), sends a constant current i through the circuit. The wire ab has a length l and mass m and can slide on the smooth, horizontal rails connected to $I_{g}$ . The entire system lies in a vertical magnetic field B. Find the velocity of the wire as a function of time.

Join BYJU'S Learning Program

Net emf =E -Bvl i=E−Bvlr (from b to a) F=ilB=(E−Bvlr) lB=(lBr) (E -Bvl)towards right After some time, when E =Bvl, then the wire moves in constant velocity v. Hence v=EBl

Net emf =E -Bvl

$i = \frac{E - B v l}{r}$ (from b to a)

$F = i l B = (\frac{E - B v l}{r})$

$l B = (\frac{l B}{r})$ (E -Bvl)towards right After some time,

when E =Bvl, then the wire moves in constant velocity v.

Hence $v = \frac{E}{B l}$