Problem 4. A straight conductor carries a current 1-10.0 A out of the page. Anot

Question

Problem 4. A straight conductor carries a current 1-10.0 A out of the page. Another straight conductor carries a current 12 15.0 A left to right in the plane of the page. The shortest distance between the conductors is (a) Determine the magnitude and direction of the net magnetic field at the point P that is midway be- (b) Determine the magnitude and direction of the net magnetic field at the point Q that is 5.00 cm above c) Determine the magnitude and direction of the net magnetic field at the point R that is 5.00 cm below 20.0 cm. tween the conductors. 1. 12.

Explanation / Answer

at point P

magneitc field due to I1 , B1 = uo*I1/(2*pi*r1) = 4*pi*10^-7*10/(2*pi*0.1) = 2*10^-5 T   ( +x direction)

magneitc field due to I2 , B2 = uo*I2/(2*pi*r2) = 4*pi*10^-7*15/(2*pi*0.1) = 3*10^-5 T ( outof page )

B = sqrt(B1^2+B2^2) = 3.6*10^-5 T

direction = tan^-1(B2/B1) = 56.3

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(b)

at point Q

r1 = 5 cm = 0.05 m

r2 = 25 cm = 0.25 m

magneitc field due to I1 , B1 = uo*I1/(2*pi*r1) = 4*pi*10^-7*10/(2*pi*0.05) = 4*10^-5 T   ( -x direction)

magneitc field due to I2 , B2 = uo*I2/(2*pi*r2) = 4*pi*10^-7*15/(2*pi*0.25) = 1.2*10^-5 T ( outof page )

B = sqrt(B1^2+B2^2) = 4.18*10^-5 T

direction = tan^-1(B2/B1) = 16.7

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(c)

at point R

r1 = 25 cm = 0.25 m

r2 = 5 cm = 0.05 m

magneitc field due to I1 , B1 = uo*I1/(2*pi*r1) = 4*pi*10^-7*10/(2*pi*0.25) = 0.8*10^-5 T   ( +x direction)

magneitc field due to I2 , B2 = uo*I2/(2*pi*r2) = 4*pi*10^-7*15/(2*pi*0.05) = 6*10^-5 T ( into the page )

B = sqrt(B1^2+B2^2) = 6.05*10^-5 T

direction = tan^-1(B2/B1) = 82.4

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