A long straight wire carriers a current i1 = 10 A. A rectangular loop carrying a

Question

A long straight wire carriers a current i1 = 10 A. A rectangular loop carrying a current of i2 = 5 A is on the right side of it. Given d = 10cm, Ldc = 12cm, Ldc = 18 cm , please calculate the magnetic field B-1, including its direction, due to i1 at where the segment ab of the loop is; the magnetic field B-2, including its direction, due to i1, at where the segment dc of the loop is; F -ab including its direction, which is the force B-1 acting on the segment ab of the loop; F-ad including its direction, which is the force B-2 acting on the segment dc of the loop; F-ad including its direction, which is the force acting on the segment ad of the loop by the magnetic field due to i1. Hint: use integration since the magnetic field is varying along ad.

Explanation / Answer

a) B1 = mue*I1/(2*pi*d)

= 4*pi*10^-7*10/(2*pi*0.1)

= 2*10^-5 T (in to the page)

b) B2 = mue*I1/(2*pi*(d+bc))

= 4*pi*10^-7*10/(2*pi*(0.1+0.12))

= 9.1*10^-6 T (in to the page)

c) Fab = B1*I2*Lab*sin(90)

= 2*10^-5*5*0.18

= 1.8*10^-4 N (towards right)

d) Fcd = B2*I2*Lcd*sin(90)

= 9.1*10^-6*5*0.18

= 8.28*10^-6 N(towards left)

e) Fad = inegral (mue*I1/2*pi*r)*I2*dr

= mue*I1*I2/(2*pi)*ln( (0.1+0.12)/0.1 )

= 4*pi*10^-7*10*5/(2*pi)*ln( 0.22/0.1)

= 7.88*10^-6 N

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