Question

Figure shows a long wire bent at the middle to form a right angle. Show that the magnitudes of the magnetic fields at the point P, Q, R and S are equal and find this magnitude.

Answer 1

The magnitudes of the magnetic fields at the point P, Q, R and S are equal and it is nearly equal to $\frac{\mu_{0} i}{2 \pi d}$

Explanation:

Step 1:

As in the figure, points P, Q, R, and S lie on a sphere with the d-radius.

Let's call cables  and .  

The magnetic field because of the $W_1$ wire is given by point P

$B_1$ = 0

The magnetic field because of the $W_2$ wire is given by point P

$B_{2}=\frac{\mu_{0} i}{2 \pi d}$  (Standing parallel to the plane in the direction outwards)

$B_{\text {net }}=\frac{\mu_{0} i}{2 \pi d}$  (Standing parallel to the plane in the direction outwards)

Step 2:

The magnetic field caused by wire $W_1$ is given by point Q

$B_{1}=\frac{\mu_{0} i}{2 \pi d}$  (Standing parallel to the plane in the direction inwards)

The magnetic field caused by wire $W_2$ is given by point Q

$B_2$= 0

$B_{\text {net }}=\frac{\mu_{0} i}{2 \pi d}$   (Standing parallel to the plane in the direction inwards)

Step 3:

The magnetic field caused by wire $W_1$ is given by point R

$B_1$ = 0

The magnetic field caused by wire $W_2$ is given by point R

$B_{2}=\frac{\mu_{0} i}{2 \pi d}$   (Standing parallel to the plane in the direction inwards)

$B_{\text {net }}=\frac{\mu_{0} i}{2 \pi d}$  (Standing parallel to the plane in the direction inwards)

Step 4:

The magnetic field caused by wire $W_1$ is given by point S

$B_{1}=\frac{\mu_{0} i}{2 \pi d}$   (Standing parallel to the plane in the direction outwards)

The magnetic field because of the $W_2$ wire is given by point S

$B_2$ = 0

$B_{\text {net }}=\frac{\mu_{0} i}{2 \pi d}$  (Standing right angles to the plane in the direction outwards)

The magnetic field magnitude at points P, Q, R and S is therefore  $\frac{\mu_{0} \mathrm{i}}{2 \pi d}$