A square loop ABCD of side a is formed by connecting wires. The resistance of the path ABC is R and the resistance of the path ADC is 2R . A steady current I₀ enters the loop at corner A and leaves at the diagonally opposite corner C. The magnitude of the net magnetic field at the center of the square loop is:

Step-by-step solution

The total current I₀ splits at corner A into two parallel paths: ABC and ADC. Given the resistances R_ ABC = R and R_ ADC = 2R , the currents in the branches are inversely proportional to their resistances. I_ ABC = 2R R + 2R I₀ = 2I₀ 3 I_ ADC = R R + 2R I₀ = I₀ 3 The magnetic field at the center of a square of side a due to one of its sides carrying current I is: B_ side = ₀ I 4 (a/2) ( 45^ + 45^ ) = ₀ I 2 a ( 2 2 ) = 2 ₀ I a Path ABC consists of two sides (AB and BC). By the right-hand rule, both produce a magnet