Question

A long straight wire parallel to the y-axis carries a current of I1 = 3.6 A in the -y-direction crosses the x-axis at a point that is 12.0 cm to the left of the origin. Another long straight wire that is 8.40 cm to the right of the origin on the x-axis carries a current of I2 = 4.2 into the page.

a) What is the net magnetic field (magnitude and direction) produced by these currents at the origin?

b) A proton moves through the origin traveling along the x-axis to the right. At the moment the proton is at the origin, what is the magnetic force on the proton?

Answer 1

dy Y3.6A CII 2 Х K 12am 8 40cm 2=4.2A long wire we know magnetic field due to Mol 20d so, At origin (we need to do vector sum) due to wire a Mol, 217 (12x103) due to wire Mo I (%) 21708.40X1002) x 4.2 Net magnetic field A 457 x 10 ? x 3.6 () + 47 X10-17 ) 27 12x10-2) 26 (8.40X10) (0.68109 () + 1615 (5))T magnetic fance on proton ti 8 - 32 magnetic bag (+8) just need to put value of CS V Scanned with CamScander His is the required Solution