Question
A rectangular conductive loop is brought in the vicirity of a straight conductor. The currents I1 = 1 A and I2 = 2 A flow it the directions shown, through the straight conductor and the rectangular loop, respectively. What is the magnitude aad direction of the net force acting on the loop? Without doing any calculation specify what to the direction and the magnitude of the force on the straight conductor. (Explain your reasoning). What is the magnitude and direction of the magnetic field at point P. at the center of the loop, produced by the straight conductor? What to the magnetic moment of the loop and in which direction it is oriented?
Explanation / Answer
Given: Distance between st. conductor to nearby edge of recangualar conductive loop = r1 = 1 m lly, distance from between st .conductor to outer edge of the loop is : r2 = 1 + 2 = 3 m currents passing through st. conductor = i1 = 1A currents passing through loop = i2 = 2A Let , F1 be the force on nearby edge of the loop : = [o i1 i2 /2r1 ] D Let , F1 be the force on outer edge of the loop : = [ o i1 i2 /2r2 ] D Note: force on the parallel branch of the loop is cancelled out. thus, Net force acting : F = F1 - F2 = [o i1 i2 D/2 ] [1/ r1 - 1/ r2] = 10x10-7 (1)(2) (4) [1/1 - 1/3 ] = 13.33x10-7 N Direction: away from the current carrying st .conductor plz. post remaining posts seperetely.., Distance between st. conductor to nearby edge of recangualar conductive loop = r1 = 1 m lly, distance from between st .conductor to outer edge of the loop is : r2 = 1 + 2 = 3 m currents passing through st. conductor = i1 = 1A currents passing through loop = i2 = 2A Let , F1 be the force on nearby edge of the loop : = [o i1 i2 /2r1 ] D Let , F1 be the force on outer edge of the loop : = [ o i1 i2 /2r2 ] D Note: force on the parallel branch of the loop is cancelled out. thus, Net force acting : F = F1 - F2 = [o i1 i2 D/2 ] [1/ r1 - 1/ r2] = 10x10-7 (1)(2) (4) [1/1 - 1/3 ] = 13.33x10-7 N Direction: away from the current carrying st .conductor plz. post remaining posts seperetely.., Note: force on the parallel branch of the loop is cancelled out. thus, Net force acting : F = F1 - F2 = [o i1 i2 D/2 ] [1/ r1 - 1/ r2] = 10x10-7 (1)(2) (4) [1/1 - 1/3 ] = 13.33x10-7 N Direction: away from the current carrying st .conductor plz. post remaining posts seperetely..,