Figure a shows, in cross section, two long, parallel wires carrying current and

Question

Figure a shows, in cross section, two long, parallel wires carrying current and separated by distance L. The ratio i_1/i_2 of their currents is 6.30; the directions of the currents are not indicated. Figure b shows the y component B_y of their net magnetic field along the x axis to the right of wire 2. The vertical scale is set by B_ys = 5.4 nT, and the horizontal scale is set by x_s = 21.9 cm. At what value of x > 0 is B_y maximum? If i_2 = 3.03 mA, what is the value of that maximum? What is the direction (into or out of the page) of i_1 and i_2?

Explanation / Answer

From the curve

At x = xs/2 = 21.9 /2 cm = 10.95cm , By = 0

So, at x = 10.95 m, Bnet = u*I1/(2*pi*(x+L)) - u*I2/(2*pi*x) = 0

So, I1/I2 = (x+L)/x = 6.3

So, (10.95+L)/10.95 = 6.3

So, L = 58.04 cm

At x = xs = 21.9 cm,

Bnet = u*I1/(2*pi*(xs + L)) - u*I2/(2*pi*xs) = (5.4/2)*10^-9

=> 6.3*u*I2/(2*pi*(xs + L)) - u*I2/(2*pi*xs) =  (5.4/2)*10^-9

=> (u*I2/(2*pi))*(6.3/(xs +L) - 1/xs) = 2.7*10^-9

=> (1.26*10^-6*I2/(2*pi))*(6.3/(0.219 + 0.5804) - 1/0.219) = 2.7*10^-9

So, I2 = 4.06*10^-3 A <------- into the page

and I1 = 6.3*I2 = 2.56*10^-2 A <------ out of the page

For Bnet to be maximum,

d(Bnet)/dx = 0

So, d/dx *(u*I1/(2*pi*(x + L)) - u*I2/(2*pi*x) ) = 0

So, -6.3*I2/(x+L)^2 + I2/x^2 = 0

So, ((x+L)/x)^2 = 6.3

So, ((x+58.04)/x) = 2.51

So, x = 38.4 cm <--------answer (a)

So, the maximu value :

Bnet,max = 6.3*u*I2/(2*pi*(x + L)) - u*I2/(2*pi*x)

= 6.3*1.26*10^-6*3.03*10^-3/(2*pi*(0.384+0.5804)) - 1.26*10^-6*3.03*10^-3/(2*pi*0.384)

= 2.39*10^-9 T

= 2.39 nT <-----answer (b)

Direction of I1 = out of the page

Direction of I2 = into the page

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