Physics (magnetic) Homework due

AI Thread Summary
A wire with a linear mass of 1 g/cm is placed on a surface with a kinetic friction coefficient of 0.2 and carries a current of 1.5 Amps. The discussion focuses on determining the smallest magnetic field required for the wire to move horizontally to the North while the current flows East. Key forces acting on the wire include its weight, normal force, Lorentz force, and friction force, which must balance for movement. The friction force is calculated as 0.2 times the weight of the wire, while the Lorentz force is expressed through the equation F = I(L x B). Understanding the relationship between the magnetic field and the current is essential for solving the problem.
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Physics (magnetic) Homework due need help urgent

I have a question. Please help

A wire having a linear mass of 1 g/cm is place on a horizontal surface that has a kinetic coefficient of friction of 0.2 . The wire carries a current of 1.5Amps toward the East, which is to the right of this page, and slide horizontally to the North , which is the top of this page. What are the magnitude and direction of the smallest magnetic field that enables the wire to move in this manner?
 
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Welcome to PF usatkling!

What are the forces acting on the body? Its weight, the normal reaction from the surface, a Lorentz Force and a friction force. Their resultant will cause the body to accelerate unless their resultant is zero.
 
I still didn't get it. It make no sense. There is no electric field involve here. The question is how to relate The external magetic field B to the current of the wire.
I know the resultant force in the moving need to be equal to the friction force
 
all u have to do is set the force of friction equal to the force equation between to magnetic fields in two wires, i can't think of the formula right now but I am sure u can find it.

the friction should be uN, which is (.2)(mg)-> (.2)((1 g/cm)*L g) where L is length of the wire
 
The Lorents Force on a conductor of length L carrying a current I when placed in a uniform magnetic field B is given by

\vec{F}=I(\vec{L}X\vec{B})
 

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