Calculating Current Needed to Lift a Cross Piece in a Horizontal Magnetic Field

In summary, the conversation discusses a physics homework problem involving a vertical H-shaped device with a sliding cross piece. The device is placed in a horizontal magnetic field that varies with the vertical distance from the bottom of the H. The question asks for the required current to lift the cross piece to a height of 0.05m. It is determined that the problem involves magnetism and that gravity must be taken into account in order to find the equilibrium current. Additionally, it is noted that the magnetic field will always be zero at the bottom, making it impossible for the cross piece to move from the bottom. The question is then clarified to ask for the current that will hold the cross piece at a height of 0.05m.
  • #1
Physicsisfun2005
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I been having trouble lately with my phyiscs HW probs. Here is only that ti'm confused on:

Three wires are connected to form a vertical H. The cross piece, mass 0.050 kg and length 0.10m , can slide up and down the vertical wires without friction. The device is placed in a horizontal field that varies B=0.10y where y is the vertical distance from the bottom of the H. What current in the H is required to lift the cross piece .05m


MY q's:

Is this a magnetism prob? (thats what we are learning in class) also can i make the cross piece effectively start from the bottom? Must I incorporate gravity?
 
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  • #2
Hi,

Yeah, it's magnetism allright. A magnetic field will exert a force on a current carrying wire. If you look at the H from the front and the current is going from left to right and the magnetic field points 'into' the H (away from you) then the magnetic force on the wire will point upward.
If you don't incorporate gravity, the wire wil simply accelerate according to Newton's second law. You'll have to find the right current for equilibrium.

Since the B-field increases as y increases, I don't think you have to start from the bottom or else the piece will just keep accelerating.
 
  • #3
I see the same problem that Galileo sees:

The question ought to be "what current will hold the cross-piece at .05 m." For this you just need to balance two equations, one for the force of gravity, and one for the Lorenz force (that's the "BIL" formula). Solve for I.

Accoring to "B = 0.10y" the magnetic field will always be zero at the bottom, so NO current will ever get it to move from the bottom since B=0. If you did give the cross-piece a little "bump," then any current that lifts it at all will cause a greater and greater force and therefore a greater acceleration as it rises. Double check the way the question is phrased.
 

1. How do you calculate the current needed to lift a cross piece in a horizontal magnetic field?

To calculate the current needed, you can use the formula I = (mg)/(Bsinθ), where I is the current, m is the mass of the object, g is the acceleration due to gravity, B is the strength of the magnetic field, and θ is the angle between the magnetic field and the direction of motion.

2. What units should be used when calculating the current?

The units for current are amperes (A).

3. What is the direction of the current in this scenario?

The direction of the current will be perpendicular to both the magnetic field and the direction of motion.

4. Can this formula be used for any object in a horizontal magnetic field?

Yes, this formula can be used for any object as long as the magnetic field is horizontal and the object is free to move.

5. How does the mass of the object affect the current needed?

The greater the mass of the object, the greater the current needed to lift it in the horizontal magnetic field. This is because the force of gravity (mg) increases with mass, and therefore the current must also increase to counteract this force.

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