\"A Chapter 19 Homework To Long Parallel Wires X X > C www.webassigrunet/web/Stu

Question

"A Chapter 19 Homework To Long Parallel Wires X X > C www.webassigrunet/web/Student/Assgrnment-Responses last?dep-17916646 ShoeDazzle Shoppin Appen Global Apps Other bookmark In the figure below, the current in the long, straight wire is 11-6.50 A, and the wire lies in the plane of the rectangular loop, which carries 10.4 A. The dimensions shown are 0.100 m,0.150 m, and0.450 m. Find the magnitude and direction of the net force exerted by the magnetic field due to the straight wire on the loop magnitude directionSelect

Explanation / Answer

Given

I1 = 6.50 A

I2 = 10.4 A

c = 0.10 m, a = 0.15 m , l = 0.450 m

here the magnetic field on straight wire due to current I1 at c is

B1 = mue0*I1/(2pi*c)

exerts a force on wire of length l , in the loop through which the current flowing is I2 is

F1 = B1*I2*l sin90

F1 = (mue0*I1/(2pi*c))*I2*l sin90

F1 = (4pi*10^-7*6.5*10.4*0.45)/(2pi*0.10) N

F1 = 6.084*10^-5 N

and due to the other length of wire l which is at a distance of (a+c) distance

from straight wire (I1) ,

the magnetic field on straight wire due to current I1 at (a+c) is B2

B2 = mue0*I1/(2pi*(a+c))

F2 = B2*I2*l sin90

F2 = (mue0*I1/(2pi*(a+c)))*I2*l sin90

F2 = (4pi*10^-7*6.5*10.4*0.45)/(2pi*0.125) N

F2 = 4.8672*10^-5 N  

so the net force is F = F1-F2

F = 6.084*10^-5-4.8672*10^-5 N

F = 1.2168*10^-5 N

the direction is to the right

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