Question

Two long straight wires, carrying currents I1 ​ and I2 ​, are placed perpendicular to each other, of shortest distance between them. PQ, has length r. The magnitude of the magnetic field at the O of PQ is A current of field at the centre 0 of pq

Answer 1

The final answer is $\frac{\mu_0}{\pi r} \sqrt{I_1^2 + I_2^2}$

Answer 2

The problem describes two long straight wires carrying currents $I_1$ and $I_2$. They are placed perpendicular to each other, and the shortest distance between them is a segment PQ of length $r$. We need to find the magnitude of the magnetic field at point O, which is the center of PQ.

1. Set up the Coordinate System:

Let's assume Wire 1 lies along the x-axis, carrying current $I_1$ in the +x direction. So, its equation is $y=0, z=0$.

Since the wires are perpendicular and have a shortest distance $r$, they must be skew lines. Let Wire 2 lie parallel to the y-axis, shifted in the z-direction. So, its equation is $x=0, z=r$. It carries current $I_2$ in the +y direction.

The shortest distance between these two wires is the line segment connecting the origin $(0,0,0)$ on Wire 1 to the point $(0,0,r)$ on Wire 2. This segment lies along the z-axis and has length $r$. Let's call these points P and Q, respectively.

P = $(0,0,0)$ Q = $(0,0,r)$

The center O of PQ is the midpoint of this segment:

O = $\left(\frac{0+0}{2}, \frac{0+0}{2}, \frac{0+r}{2}\right) = (0,0, r/2)$.

2. Magnetic Field due to Wire 1 ($B_1$):

Wire 1 is along the x-axis ($y=0, z=0$), carrying current $I_1$ in the +x direction.

Point O is at $(0,0, r/2)$.

The perpendicular distance from Wire 1 to point O is $d_1 = \sqrt{(0-0)^2 + (0-0)^2 + (r/2-0)^2} = r/2$.

The magnitude of the magnetic field due to a long straight wire is $B = \frac{\mu_0 I}{2\pi d}$.

So, $B_1 = \frac{\mu_0 I_1}{2\pi (r/2)} = \frac{\mu_0 I_1}{\pi r}$.

Using the right-hand rule (thumb in +x direction, fingers curl), at a point $(0,0,z)$ with $z>0$, the magnetic field points in the +y direction.

Therefore, $\vec{B_1} = \frac{\mu_0 I_1}{\pi r} \hat{j}$.

3. Magnetic Field due to Wire 2 ($B_2$):

Wire 2 is along the y-axis at $z=r$ ($x=0, z=r$), carrying current $I_2$ in the +y direction.

Point O is at $(0,0, r/2)$.

The perpendicular distance from Wire 2 to point O is $d_2 = \sqrt{(0-0)^2 + (0-0)^2 + (r/2-r)^2} = \sqrt{(-r/2)^2} = r/2$.

The magnitude of the magnetic field due to Wire 2 is $B_2 = \frac{\mu_0 I_2}{2\pi (r/2)} = \frac{\mu_0 I_2}{\pi r}$.

To find the direction of $\vec{B_2}$, consider Wire 2 to be along the y-axis (current in +y). Point O is at $x=0, z=r/2$ relative to the origin of the coordinate system. Relative to the wire (which is at $z=r$), the z-coordinate of O is $z_{rel} = r/2 - r = -r/2$. The x-coordinate is $x_{rel} = 0$.

Using the right-hand rule (thumb in +y direction), at a point $(x,z)$ relative to the wire:

If current is in +y, the magnetic field is in the x-z plane. For $x_{rel}=0, z_{rel}=-r/2$, the point O is "below" the wire in the x-z plane. The field direction will be in the $-\hat{i}$ direction.

Alternatively, for a wire along the y-axis, $\vec{B} \propto (z \hat{i} - x \hat{k})$. For Wire 2, $x_{rel}=0, z_{rel}=-r/2$.

So, the direction is $(-r/2)\hat{i} - 0\hat{k} = -r/2 \hat{i}$. This is in the $-\hat{i}$ direction.

Therefore, $\vec{B_2} = \frac{\mu_0 I_2}{\pi r} (-\hat{i})$.

4. Total Magnetic Field at O:

The total magnetic field at O is the vector sum of $\vec{B_1}$ and $\vec{B_2}$:

$\vec{B} = \vec{B_1} + \vec{B_2} = \frac{\mu_0 I_1}{\pi r} \hat{j} - \frac{\mu_0 I_2}{\pi r} \hat{i}$

$\vec{B} = \frac{\mu_0}{\pi r} (-I_2 \hat{i} + I_1 \hat{j})$

5. Magnitude of the Total Magnetic Field:

The magnitude of $\vec{B}$ is $|\vec{B}| = \sqrt{B_x^2 + B_y^2 + B_z^2}$.

$|\vec{B}| = \sqrt{\left(-\frac{\mu_0 I_2}{\pi r}\right)^2 + \left(\frac{\mu_0 I_1}{\pi r}\right)^2 + 0^2}$

$|\vec{B}| = \sqrt{\frac{\mu_0^2}{\pi^2 r^2} (I_2^2 + I_1^2)}$

$|\vec{B}| = \frac{\mu_0}{\pi r} \sqrt{I_1^2 + I_2^2}$