(Here the position of wires and point P is not specified. So i have assumed the the wire 1 is on the left, wire 2 on right and point p on right to both of them)
Suppose the current in second wire is I2. Distance from the second wire is, d2 = 1.5-0.75=0.75cm = 0.0075m
d1 = 1.5cm =0.015m
So Magnetic field at P is
B =μ0I1/2πd1 + μ0I2/2μd2
=(μ0/2π)(I1/d1 + I2/d2)
Given, B = 0
=> I1/d1 + I2/d2 = 0
=> I2 = - I1×d2/d1 = -6.5×0.0075/0.015
=-3.25A
Therefore the current in second wire is 3.25A coming out of the page.
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