Three long parallel wires each carry 2.0-a currents in the same direction. the wires are oriented vertically, and they pass through three of the corners of a horizontal square of side 4.0 cm. what is the magnitude of the magnetic field at the fourth (unoccupied) corner of the square due to these wires? (μ0 = × 10-7 t · m/a)

T

Explanation:

 = magnitude of current in each wire = 2.0 A

 = length of the side of the square = 4 cm = 0.04 m

 = length of the diagonal of the square = a = (0.04) = 0.057 m

= magnitude of magnetic field by wires at A and C

T

= magnitude of magnetic field by wire at B

T

Net magnitude of the magnetic field at D is given as

T

tbh i dont know ether

the diagram which shows forces as either balanced = same speed or unbalanced = accelerarting