A long wire carrying a 5.5 A current perpendicular to the xy-plane intersects the x-axis at x=−2.0cm . A second, parallel wire c

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A long wire carrying a 5.5 A current perpendicular to the xy-plane intersects the x-axis at x=−2.0cm . A second, parallel wire c

arrying a 2.5 A current intersects the x-axis at x=+2.0cm . Part A At what point on the x-axis is the magnetic field zero if the two currents are in the same direction? Express your answer in centimeters. x x = nothing cm SubmitRequest Answer Part B At what point on the x-axis is the magnetic field zero if the two currents are in opposite directions? Express your answer in centimeters. x x = nothing cm SubmitRequest Answer Provide Feedback Next

Physics

1 answer:

Svetach [21] · Answer 1 · 2022-04-17T05:46:23+03:00

Answer:

a

The point on the x-axis where the magnetic field will be zero is $d = 2.8 \ cm$

b

The point on the x-axis where the magnetic field will be zero is $Z = 5.333 \ cm$ Explanation:

From the question we are told that

The first current is $I_1 = 5.5 \ A$

The point of x- axis intersection is $x = -2.0 \ cm$

The second current is $I_2 = 2.5 \ A$

The point of intersection of the x-axis is $x = 2 .0 \ cm$

Generally given that the current of the two wires are same direction it means that the magnetic field in -between the wires will cancel out giving zero

So

$B_1 - B_2 = 0$

=> $B_1 = B_2$

=> $\frac{\mu_o * I_1 }{2 * \pi * d } = \frac{\mu_o * I_2 }{2 * \pi * (4 -d) }$

Here d is the of one wire to the point where the magnetic field is 0

and given that the total distance in-between the wire is D = 2 = 2 = 4

Hence the distance of the other wire to the point where magnetic field is zero is (4 - d)

So

$\frac{5.5}{ d} = \frac{2.5}{4-d}$

=> $8d = 22$

=> $d = 2.8 \ cm$

So the point on the x-axis where the magnetic field will be zero is $d = 2.8$

Generally given that the current of the two wires are opposition direction it means that the magnetic field at a position which is not in-between the wire will be zero

Let that position be k

Let the distance from the middle of both wires to k be Z

So

$\frac{\mu_o * I_1 }{2 * \pi * ( 2 + Z) } = \frac{\mu_o * I_2 }{2 * \pi * (Z -2) }$