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8 Jul 2008 01:13:10 IST

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H.C.Verma chapter magnetic field questio-22 and 27 of the exercise not able to solve.

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H.C.Verma chapter magnetic field questio-22 and 27 of the exercise not able to solve.

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ramyani chakrabarty

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8 Jul 2008 01:57:55 IST

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Me too can't solve Q.22. The Q is

2 metal strips , each of length l, r clamped || to each other on a horizontal floor with a sepn of b b/w them. A wire of mass m lies on them perpendicularly as shown in fig ( see hcv vol II , page 232 ). A vertically upward magnetic field B exists in the space. The metal strips r smooth but the coeff of friction b/w the wire and the floor is n. A current i is established when the switch S is closed at the instant t = 0. Discuss the motion of the wire after the switch is closed.

How far away from the strips will the wire reach ?

anchit saini

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8 Jul 2008 07:58:44 IST

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This is what i could make out of the qn ->

The force is towards right

The wire moves from the left end to the right end covering a distance l , during which time B acts on it , thereafter it moves on the floor where friction acts on it , and at a distance x from the strips the wire stops

The energy gained due to B is = Fl = ibB l = mg x

which gives

x= $\frac{ilbB}{\mu mg}$

ramyani chakrabarty

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8 Jul 2008 10:29:14 IST

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@ anchit

I don't understand. Plz .

Conjurer

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8 Jul 2008 14:45:27 IST

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Change in KE = Work done by all the forces.

0 = Work which is apparently done by Magnetic force ( <<-- I wish this worries the kids :D)+ Work done by friction

This is what he has done.

Also see this if you dont understand : http://www.goiit.com/posts/list/magnetism-magnetic-force-work-done-62225.htm#311671

ramyani chakrabarty

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9 Jul 2008 00:20:09 IST

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Change in KE = Work done by all the forces.

0 = ibB l - mg x

x = $\frac{ilbB}{\mu mg}$

ok.

isn't there any other method ?

ramyani chakrabarty

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9 Jul 2008 01:40:39 IST

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experts plz step in.

Krishna Gopal Singh

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9 Jul 2008 09:24:08 IST

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anchit is right. See when the wire is on the strip there is no friction and magnetic force will make it move with constant acceleration gaining kinetic energy = F*l.
After it leaves the strip there is opposing force of friction which will deacelerate it loosing K.E = mu*mg*x
Equating the two we will get x.

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