Say you have a 5 cm current wire carrying 10 A going from left to right. Directly 1 c.m below the left end of this wire is a long wire that is perpendicular to the first wire and goes out of the page. What is the net force on the 5 cm wire?I've tried using F=ILB with the I of the first wire and the B of the second wire.
I didn't read the question so I was carefully working out the force. The wire is perpendicular to the first wire, so using the right hand rule you discover that the field it creates is PARALLEL to the first wire at this left end. The magnetic force is caused by the component which is PERPENDICULAR to the wire which is in fact zero. So there is no magnetic force at this point. As you move along the wire you get a diminishing amount of magnetism caused by the wire which is going out of the page but that field has a component which is DOWN the page. Therefore that part of the wire experiences a force which is into the page. ( take your right hand, put the thumb along the wire pointing to the right, the fingers point down the page, the palm points into the page which is then the direction of the force) I would be surprised if you were required to work out the magnitude of the force in this context. You can't use F= ILB because both the magnitude and the direction of the field varies at different points along the wire. If the perpendicular wire had been directly below the middle of the other wire there would have been no net force. If you were of a level where working out the force was appropriate you would need to set up the formula for B at various points along the wire, taking the vertical component only and integrate this over the range from 0 to 5 cm. Not a trivial mathematical task.
