Magnetic field due to infinite current carrying wire in the X and Y axes

• wcjy
In summary, the conversation discusses the calculation of the magnetic field in the (i-y) direction using the Right-hand Grip Rule and the distance of a point to the Y and X axis. The correct expression for the magnetic field is given as B = (5μ0I)/(82π) (i-j), and the mistake of adding the two B-fields is pointed out, with the correct method of adding vectors in a cartesian coordinate system explained. The calculation for the k-direction is not necessary as it cancels out with the other wire.
wcjy
Homework Statement
Consider 2 identical infinite current-carrying wires which runs along the X and Y axes
respectively. They carry identical currents I and the currents flow in the directions of the
axes. Find the net magnetic field (in vector form) at the coordinate (x, y, z) = (4,4,5).
Relevant Equations
Biot savar law
$$B = \frac {\mu_0 I}{2 \pi r}$$

By Right-hand Grip Rule, the direction of the magnetic field by wire in y-axis is into the paper (z)
while the direction of the magnetic field by wire in X-axis is upwards (+i)

The answer state the Magnetic field is in the (i - y) direction though.

Next calculating the magnitude,
Distance of point to the Y axis is $$\sqrt{4^2 + 4^2} = \sqrt{32}$$
Distance of point to the X axis is $$\sqrt{4^2 + 5^2} = \sqrt{41}$$

Therefore, Magnetic field = $$B = \frac {\mu_0 I}{2 \pi \sqrt{32}} +\frac {\mu_0 I}{2 \pi \sqrt{41}}$$
This will give some weird fraction which is wrong.

Correct answer is $$B = \frac {5\mu_0 I}{82 \pi } (\hat{i} - \hat{j})$$

Delta2
Oh wait is the Magnetic field from wire in Y axis in the (z and x direction) and b field from wire in x-axis in the ( -z and direction) so when adding those B field, the z cancels out?

I think so too , the z-component cancels. To see it clearly you must work with both fields in the same cartesian coordinate system. The popular expression $$\vec{B}=\frac{\mu_0}{2\pi}\frac{I}{r}\hat\phi$$ is in cylindrical coordinate system, not in cartesian.

I think in the OP, you do two mistakes: one mistake is that the distance of the (4,4,5) point to the y-axis is i believe ##\sqrt{4^2+5^2}##. The other mistake is that you can't just add those two B-fields cause each has its own ##\hat\phi## direction (hard to explain this without doing a diagram)...

wcjy
Ok i think i realized my first mistake.
Ok i just realized this. can help me check if what i said is correct.

I drew a 2d diagram for z and x-axis for the wire (yaxis).
the b field will be pointing in the i and z direction.
z direction is $$B = \frac{\mu_o I}{2 \pi r }sin \theta$$
while the i direction is $$\frac{\mu_o I}{2 \pi r }cos \theta$$
cos theta = 5/ sqrt 41
so $$B = \frac{5\mu_o I}{2 *\sqrt{41}*\sqrt{41}*\pi }$$
$$B = \frac{5\mu_o I}{82 \pi } (\hat{i})$$

for the other wire is the same.
but $$B = \frac{5\mu_o I}{82 \pi } (\hat{j})$$

But arent I suppose to add or do the sqrt( B^2 + B^2 ) thing?

Yes you are supposed to add them but add them as vectors of a cartesian coordinate system, since the answer given is a vector in a cartesian coordinate system (it contains the unit vectors ##\hat i,\hat j##).

wcjy
I think you calculated correctly the i-component of the B-field from the wire in y-axis, so the result in vector form would be ##\vec{B_{y wire}}=\frac{5\mu_0 I}{82\pi}\hat i+...\hat k## but we are missing the z or ##\hat k##-component.

Then write a similar expression for the B-field from the wire in x-axis. Then just add them as vectors.

wcjy
to 'add' them, isn't it $$\sqrt {2 (\frac{5\mu_o I}{82 \pi })^2}$$

Delta2 said:
I think you calculated correctly the i-component of the B-field from the wire in y-axis, so the result in vector form would be ##\vec{B_{y wire}}=\frac{5\mu_0 I}{82\pi}\hat i+...\hat k## but we are missing the z or ##\hat k##-component.

Then write a similar expression for the B-field from the wire in x-axis. Then just add them as vectors.
for the k direction, its just sin theta. but it cancels out with the other wire. so i left it out

wcjy said:
to 'add' them, isn't it $$\sqrt {2 (\frac{5\mu_o I}{82 \pi })^2}$$
That's the magnitude of the total vector. You are being asked for the vector of total magnetic field (if i judge from the given answer) and not for the magnitude of this vector.

wcjy
oh so if it is
$$\frac{5\mu_o I}{82 \pi } (\hat{i})$$ and $$\frac{5\mu_o I}{82 \pi } (-\hat{j})$$,
the answer is $$\frac{5\mu_o I}{82 \pi } (\hat{i}-\hat{j})$$ ??

yes that's what you get if you add the first two as we do in vector addition.

wcjy
Okay. Thanks so much!

Delta2

1. What is the direction of the magnetic field around an infinite current carrying wire in the X and Y axes?

The direction of the magnetic field around an infinite current carrying wire in the X and Y axes is circular, with the field lines forming concentric circles around the wire.

2. How is the strength of the magnetic field affected by the distance from the wire?

The strength of the magnetic field around an infinite current carrying wire in the X and Y axes decreases as the distance from the wire increases. This follows the inverse square law, meaning that the strength of the field is inversely proportional to the square of the distance from the wire.

3. Can the direction of the magnetic field be reversed by changing the direction of the current in the wire?

Yes, the direction of the magnetic field around an infinite current carrying wire in the X and Y axes can be reversed by changing the direction of the current in the wire. This is because the direction of the magnetic field is directly proportional to the direction of the current.

4. How does the magnitude of the current affect the strength of the magnetic field?

The strength of the magnetic field around an infinite current carrying wire in the X and Y axes is directly proportional to the magnitude of the current. This means that as the current increases, the strength of the magnetic field also increases.

5. Is the magnetic field uniform around an infinite current carrying wire in the X and Y axes?

No, the magnetic field around an infinite current carrying wire in the X and Y axes is not uniform. The strength of the field varies depending on the distance from the wire, with the strongest field being closest to the wire and decreasing as the distance increases.

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