Friction and Magnetism

Electricity and Magnetism Level 5

A straight rod of mass $1\text{ kg}$ and length $1\text{ m}$ can slide on two parallel plastic rails kept in a horizontal plane with separation $1\text{ cm}$ . The coefficient of friction between the wire and rails is $\sqrt3$ .

If the wire carries a current $10A$ , what minimum magnetic field would allow the wire to slide on the rails?

The answer is 0.8660254.

2 solutions

Sayantan Dhar
Aug 23, 2016

here it is given that the the rod slides on 2 plastic rails. so to find minimum force we need to slide only 1/2 of the mass. so we get umg/2=ilb. solving this we get b=0.886

Would placing infinite number of rails (making plane surface) mean no force needed? As it would mean 1/n of mass to be slided where n tends to infinity. @Sayantan Dhar

Mayank Chaturvedi - 4 years, 2 months ago

God Knows
Oct 17, 2014

actually this question is a bit tricky we know that force on a current CARRYING WIRE IS ( ILB) SO IF WE EQUATE ILB=UMG ..........................( U= COEFFICIENT OF FRICTION) WE MAY THINK THAT WE GET THE ANSWER BUT IF MAGNETIC FORCE ACTS AT AN ANGLE HOW WE DEAL WITH THAT. SO THE ACTUAL ANSWER IS DIVIDING THE THE ANSWER WHICH WE HAVE GOT ABOVE BY ............................( 1+U^2)^1/2

HINT................. FORCE ACTS AT AN ANGLE NOT DIRECTLY PERPENDICULAR

can you please explain how you got indentity..........( 1+U^2)^1/2........ for dividing.....

Pranav Patil - 6 years, 2 months ago

but minimum magnetic field is required when B is perpendicular.

Akshay Sharma - 5 years, 5 months ago

Plz explain how u got that absurd factor to divide..

Istiak Reza - 4 years, 9 months ago

YES BUDDY GOD KNOWS

Anubhav Tyagi - 4 years, 9 months ago

Friction and Magnetism

The answer is 0.8660254.

2 solutions

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