Three Straight Three infinite straight wires are fixed in place and aligned para

Question

Three Straight Three infinite straight wires are fixed in place and aligned parallel to the z-axis as shown. The Ain the negative z direction. The wine at Dx.y 11 cm,00 carries current 2 09Ain the positive z-direction. The wire at (x,y) (0, 19.1 cm) carries current ly 6.5 Ain the positive z "What is B10,0), the x-component of the magnetic field produced by these three wires at the origin? You currently have 1 submissions for this question. Only 10 submission are allowed. You can make 9 more submissions for this question. Your answer has been judged correct the exact answer is: a What is Byl0,0), the y component of the magnetic field produced by these three wires at the origin? You currently have 1 submissions for this question. Only 10 submission are allowed. You can make 9 more submissions for this question. Your submis Your answer has been judged correct: the exact answer is: -7.81818181818182 10 What is F1), thex component of the force exerted on a one meter length of the wire carrying current l? You currently have 1 submissions for this question. Only 10 submission are allowed. You can make 9 more submissions for this question. Your submitt What is Fy(11. the y component of the force exerted on a one meter length of the wire carrying current I? You currently have 1 submissions for this question. Only 10 submission are allowed. You can make 9 more submissions for this question. Your submise Your answer has been judged correct: the exact answer is: O What is F 2), the x-component of the force exerted on a one meter length of the wire carrying current

Explanation / Answer

3)

I1 = 3.4 A in negative z-direction, at (-11 cm,0)

I2 = 0.9 A in positive z-direction, at (11 cm,0)

I3 = 6.5 A in positive z-direction, at (0,19.1 cm)

Force on wire carrying current I1 due to wire carrying current I3 is,

F13 = µ0I1I3/(2d) along the line joining 1 and 3 and away from wire 3.

So component of F13 along x-axis is,

F13x = -F13cos600 = -F13/2

Force on wire carrying current I1 due to wire carrying current I2 is,

F12 = µ0I1I2/(2d) along negative x-axis,

So Fx(1) = - µ0I1I3/(4d) - µ0I1I2/(2d) = -µ0I1/(2d)[I3/2 + I2] = -[(2X10-7T-m/A)(3.4 A)/(0.22 m)][6.5 A/2 + 0.9A]

or, Fx(1) = -1.28X10-5N

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5)

Similarly as above,

Fx(2) = µ0I1I2/(2d) - µ0I3I2/(4d) = µ0I2/(2d)[I1 - I3/2]

or, Fx(2) = [(2X10-7T-m/A)(0.9 A)/(0.22m)][3.4A - 6.5A/2]

or, Fx(2) = 1.23X10-7 N

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6)

Option 1st and 2nd cannot be true as I4 and I3 will have fields at origin along or opposite of the x-axis only. So they cannot affect y-component.

Option 3rd cannot be true as, if I3 and I4 both have current along positive z-axis, then they produce field along positive x-axis at origin.

If I4 is directed along the negative z-axis, then it is possible to make the x-component of the magnetic field equal to zero at the origin.

So 4th option is correct.

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