k 9 times 109 NM2C2 Eo 885 times 1012 C2Nm2 mu o 4pi time

k = 9 times 10^9 N-M^2/C^2 Eo = 8.85 times 10^-12 C^2/N-m^2 mu o = 4pi times 10^-7 T-m/A E = 1.6 times 10^-19 C Mass electron = 9.11 times 10^-31 kg Mass proton & neutron = 1.67 times 10^-27 kg c = 3 times 10^8 m/s 1 Tesla = 1 times 10^4 Gauss Consider (infinitely long) currents I_1 = 2A and I_2 = 1A traveling in the indicated directions (each square is 1 cm). a. What is Magnetic Field at A? b. What is Magnetic Field at B?

Solution

magnetic field due to long wire B = uo*I/(2*pi*r)

at point A

r1 = 2cm = 0.02 m

r2 = 2cm = 0.02 m

magnetic field due to wire 1 , B1 = uo*I1/(2*pi*r1)   into the page

B1 = 4*pi*10^-7*2/(2*pi*0.02) = 2*10^-5 T into the page

magnetic field due to wire 2 , B2 = uo*I2/(2*pi*r2)   out of the page

B2 = 4*pi*10^-7*1/(2*pi*0.02) = 1*10^-5 T out of the page

Bnet = B1 - B2 = 1*10^-5 T into the page

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at point B

r1 = 3 cm = 0.03 m

r2 = 5 cm = 0.05 m

magnetic field due to wire 1 , B1 = uo*I1/(2*pi*r1)   out of the page

B1 = 4*pi*10^-7*2/(2*pi*0.03) = 1.33*10^-5 T out of the page

magnetic field due to wire 2 , B2 = uo*I2/(2*pi*r2)   into the page

B2 = 4*pi*10^-7*1/(2*pi*0.05) = 0.4*10^-5 T into the page

Bnet = B1 - B2 = 0.93*10^-5 T out of the page