A copper sheet with length 1.0 m, width 0.50 m, and thickness 1.0 mm is oriented so that its largest surface area is perpendicular to a magnetic field of strength 6.4 T. The sheet carries a current of 3.5 A across its length. What is the magnitude of the force on this sheet?what is the formula that involves Area when finding the Fb??can someone tell me?
Imagine the sheet made up of N thin strips of copper running lengthwise across the sheet, length L 1m, with infinitely thin insulation strips between each copper strip. The current thru each strip is 3.5A/N. The force on each strip due to the B field is F (i/N)LB pushing the strip sideways. So the total force on the sheet is N*(i/N)LB iLB, just as though the sheet were a single wire. Magnitude iLB 3.5A*1m*6.4T 22.4 N answer
Imagine the sheet made up of N thin strips of copper running lengthwise across the sheet, length L 1m, with infinitely thin insulation strips between each copper strip. The current thru each strip is 3.5A/N. The force on each strip due to the B field is F (i/N)LB pushing the strip sideways. So the total force on the sheet is N*(i/N)LB iLB, just as though the sheet were a single wire. Magnitude iLB 3.5A*1m*6.4T 22.4 N answer
