What Magnetic Field Strength Is Needed to Lift Ferritic Stainless Steel?

[sherryjenix]

Homework Statement

Hi, I am trying to calculate the current and magnetic field strength needed to pick up a 0.3125 lb. piece of a channel lock plier made of ferritic stainless steel (they have a soft magnetic component) on its y-axis (on the lighter part of the piece) at an 80 degree angle. The magnet would be located 9.36 in. above the piece of steel.

Homework Equations

Fb = ILB
Fb = qvB
B = ($$\mu$$0I)/[4pi(y)]

The Attempt at a Solution

I drew a freebody diagram and found that the magnetic force needed to pick up the piece of channel lock plier would be 12.3N. I did more research and found that a magnetic field force would only be exerted on moving charge particles.

Now I am stumped on how to find the current and magnetic field strength without knowing the other.

 

[_coyote_]

I tried to use the equation Fb=qvB, but that didn't get me anywhere either. Can anyone help?You can use the equation Fb = ILB, where Fb is the magnetic force in Newtons, I is the current in Amps and LB is the length of the plier in meters. Plugging in the values given, you get: Fb = I(0.3125 lb)(9.36 in.) Rearranging, this gives you: I = (Fb)/((0.3125 lb)(9.36 in.)) Now you know the current, you can use the equation B = (\mu0I)/[4pi(y)] to calculate the magnetic field strength. Plugging in the values given, you get: B = (\mu0I)/[4pi(9.36 in.)] Rearranging, this gives you: B = (\mu0((Fb)/((0.3125 lb)(9.36 in.))))/[4pi(9.36 in.)] Therefore, you can calculate the magnetic field strength given the magnetic force needed to pick up the piece of channel lock plier.