# How Do Antiparallel Currents in Two Wires Affect the Magnetic Field at a Point?

• xinlan
In summary, Homework Equations state that the net magnetic field at point L due to a wire with current I and distance d is B_M=2(u_oI)/(2pi*sqrt of d^2+(2d)^2)cos(63.4).
xinlan

## Homework Statement

In this problem, you will be asked to calculate the magnetic field due to a set of two wires with antiparallel currents as shown in the diagram . Each of the wires carries a current of magnitude . The current in wire 1 is directed out of the page and that in wire 2 is directed into the page. The distance between the wires is . The x-axis is perpendicular to the line connecting the wires and is equidistant from the wires.

Find the magnitude of the net magnetic field created at point L by both wires.

Bwire = uoI/2pid

## The Attempt at a Solution

I got the B1L created by wire 1 by point 1 only.
so I thought since they have the same current and distance, they will have the same magnitude, so I just add them. But I got it wrong.
one of the hint says "find the direction of B"
I am confused how to find the direction for wire 1 and 2 at point L.

Thank you so much..

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The magnetic field surrounding a current-carrying wire is tangent to a circle centered on that wire. Use the right-hand rule to find which way it points. For the current coming out of the page, point your thumb in the direction of current (out) and your fingers curl in the direction of the field (counter clockwise).

I had used the right hand rule for both currents. the other current will be clockwise right?
but I still don't understand how come it will be relevant to the question. since the hand rule only gives either counterclockwise or clockwise.

xinlan said:
I had used the right hand rule for both currents. the other current will be clockwise right?
Right.
but I still don't understand how come it will be relevant to the question. since the hand rule only gives either counterclockwise or clockwise.
Use that rule to find the direction of magnetic field from each wire. Note that the magnetic field is a vector quantity, so you need to add them like vectors. (You can't just add them like numbers.)

For example: To find the field from wire 1 at location L, draw a line from the wire to L. The field will be perpendicular to that line with magnitude given by the formula, in a counterclockwise direction. (You'll have to figure out the distance between 1 and L, of course.) Draw the vector on the diagram.

Then do the same thing for the field from wire 2. Draw its magnetic field vector at point L.

Then add these vectors to find the net field at point L. (Find their components and add them.)

2*((uo*I)/(2pi*sr of d^2+(d*sr of 2)^2)*cos55

where sr = square root

B1x = + B1 cos 55
B2x = + B2 cos 55
so bx = B1x+B2x = 2*(B*cos 55)
where B = ((uo*I)/(2pi*sr of d^2+(d*sr of 2)^2)

then, net B = sr of Bx^2
= sr of ((2(B*cos 55))^2)
= 2(B*cos 55)
= 2*((uo*I)/(2pi*sr of d^2+(d*sr of 2)^2)*cos55

What's the distance (in terms of d) between wire 1 and point L?

this is the answer that how I put in:

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xinlan said:
this is the answer that how I put in:
Simplify that expression for the distance. I think the system is too stupid to understand.

this is the answer how I type in:
but I still got wrong

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Get rid of the angle. Express the cos(55) in terms of d only.

Doc Al, thank you so much for your help..
I really appreciate it..

But i didnt get it>...<

Similar Problem

Find the magnitude of the net magnetic field B_M created at point M by both wires.

B_m=2(u_oI)/(2pi*sqrt of d^2+(2d)^2)cos(63.4) but I rewrote cos(63.4) in terms of d, (0.5) but my answer is off by a multiplicative factor.

Could you help me?

Doc, would you be so kind as to explain how we can rewrite cos(55) in terms of d. I am not quite sure how to do this.

Thanks.

fubag said:
Doc, would you be so kind as to explain how we can rewrite cos(55) in terms of d. I am not quite sure how to do this.
Consider the right triangle 1-K-L. All sides can be expressed in terms of d. The 55 degrees corresponds to the angle at point 1; write cos(55) in terms of the ratio of the sides. (The answer will not have a d in it, of course, since they cancel.)

hint remember cos(theta) is = x/r ...in this case x is d! and r is length of b!

Sorry to bump a old thread but I just had this question in mastering physics. and it Said it MUST BE in radians.. so the 55 would have to be 0.9547. and that's all that needed to be changed to make it correct

## 1. What is the magnetic field produced by two parallel wires?

The magnetic field produced by two parallel wires is determined by the distance between the wires, the current flowing through each wire, and the direction of the current. The magnetic field is strongest between the two wires and decreases as you move further away from the wires.

## 2. How does the direction of the current affect the magnetic field from two wires?

The direction of the current in each wire determines the direction of the magnetic field. If the currents are flowing in the same direction, the magnetic fields will be parallel and additive. If the currents are flowing in opposite directions, the magnetic fields will cancel each other out in certain regions.

## 3. What is the formula for calculating the magnetic field from two wires?

The formula for calculating the magnetic field from two wires is B = (μ0 * I) / (2 * π * r), where μ0 is the permeability of free space, I is the current, and r is the distance between the wires.

## 4. How does the distance between the wires affect the strength of the magnetic field?

The distance between the wires is directly proportional to the strength of the magnetic field. As the distance increases, the magnetic field decreases. This is because the field lines spread out over a larger area, resulting in a weaker magnetic field.

## 5. How can the magnetic field from two wires be used in practical applications?

The magnetic field from two wires can be used in various practical applications such as in electromagnets, electric motors, and generators. It can also be used in medical imaging techniques such as MRI machines and in particle accelerators for scientific research.

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