Two long straight parallel wires separated by a distance of 20 cm carry currents

Question

Two long straight parallel wires separated by a distance of 20 cm carry currents of 30 A and 40 A in opposite directions. What is the magnitude of the resulting magnetic field at a point that is 15 cm from the wire carrying the 30-A current and 25 cm from the other wire? (Give your answer in micro Tesla: 1 mu T =10^-6 T) A planar loop consisting of four turns of wire, each of which encloses 200 cm^2, is oriented perpendicularly to a magnetic field that increases uniformly in magnitude from 10 mT to 25 mT in a time of 5.0 ms. What is the resulting induced current in the coil if the resistance of the coil is 5.0 Ohm? (Give your answer in mA)

Explanation / Answer

1A) distance between the wire is 20cm and 30A wire to point P is 15cm and distance from 40A wire to P is 25cm

as we can see , 25^2 = 15^2 + 20^2

625 = 225 + 400 = 625

so this is a right angle traingle.

P is directly above the 30A wire and makes an angle of tan^-1(15/20) from 40A with line joining 30A and 40A.

@ = tan^-1(15 / 20) = 37 deg .

angle between magnetic field vector due to 30A and 40A will be 90+37 = 127 deg

B1 = u0I1 / 2pi r = (4pi x 10^-7 x 30 ) / (2 pi 0.15) = 40 x 10^-6 T

B1 = 40 uT

B2= (4 pi x 10^-7 x 40 ) / (2 pi 0.25 ) = 32 x 10^-6 T

B2 = 32 uT

Bnet = sqrt[ B1^2 + B2^2 + 2B1B2cos127 ]

= 32.91 uT

1b)

induced emf = rate of change of flux = d(NBA)/dt

here field is changing with time, dB/dt = (25 - 10) mT / 5ms = 3 T/s

e = N A dB/dt

= 4 x 200 x 10^-4 x 3 =0.24 V

I = e/R = 0.24 / 5 =0.048 A = 48 mA

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