Consider a straight wire that has a mass per length of 100 g/m, a fixed charge p

Question

Consider a straight wire that has a mass per length of 100 g/m, a fixed charge per unit length of ?35 ?C/m, and a current can run from left to right as shown on the attached figure. A uniform electric field of 12000 N/C exists that is directed straight up, and there is a uniform B field of magnitude 0.04 T that is free to point in any direction.

a) Determine the direction of the magnetic field such that the wire could be suspended above the ground.

b) Determine the magnitude of the current such that the wire will be suspended above the ground.

Explanation / Answer

a)

For equilibrium,

net force = 0

forces acting on the wire :

1) force of gravity = mg <--- acting downwards

2) magnetic force = BIL <--- must act in a direction such that net force = 0

3) electric force = Q*E <------ acting downwards

So, BIL = mg - EQ

So, 0.4*I = (m/L)*g + E*(Q/L)

So, 0.4*I = (0.1)*9.8 + 12000*35*10^-6

So, I = 3.5 A <--------- current needed (ANSWER)

Now, for magnetic force to be directed in the upwards direction, the magnetic field must be directed into the page

So, answer is into the page <----------answer(for magnetic field )

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