Two long straight parallel wires carry 5.0 A and 8.0 A currents in opposite dire

Question

Two long straight parallel wires carry 5.0 A and 8.0 A currents in opposite direction as shown. The separation between the wires is 10.0 cm. (a) Determine the force per unit length on each wire, identifying if it is attractive or repulsive, (b) Determine the magnitude and direction of the magnetic field at: (i) P, the midpoint between the wires, (ii) A, a point 10.0 cm to the left of the wire that carries 5.0 A, (iii) B, a point 10.0 cm to the right of wire that carries 8.0 A current. Indicate the direction of the field in each case.

Explanation / Answer

(a) We know that when the current flows in the opposite direction then they have repulsive force.
F = k*2(I1I2)/r where k is constant = 10-7
F = 10-7*(2*5*8)/(10*10-2) = 8*10-5 N
(b) We know that the magnetic field due to the infinite wire is
B = k(2I/r)
where r is the perpendicular distance of the point from the wire.
(i) With the help of the right hand screw rule we can find the direction of the magnetic field. Hence the direction of the magnetic field because of both wire will be in the same direction that is inside the paper.
B = B1 + B2 = k(2/r)(I1+ I2) = 10-7*(2/0.1)(5+8) = 2.6*10-5 T
(ii) At point A the magnetic field due to wire carrying 5 A current will be outside of the paper and due to 8 A current will be inside the paper
B = B1 -B2 (considering outside coming magnetic field as positive)
B = 2k(I1 /r1 - I2/r2 ) = 2*10-7 ((5/0.1) - 8/0.3) = 4.67*10-6 T
(iii) In this case the direction of the magnetic field due to 5 A wire will be inside the paper and 8 A will be outside the paper therefore
B = B2 -B1 = 2k( I2/r2 - I1 /r1 ) = 2*10-7 ((8/0.1) - (5/0.3)) = 1.267*10-5 T

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