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

As shown in the figure, points P, Q, R and S lie on a circle of radius d.
Let the wires be named W₁ and W₂.


Now,
At point P, the magnetic field due to wire W₁ is given by
B₁ = 0
At point P, the magnetic field due to wire W₂ is given by
$B_2=\frac{{\mathrm{\mu }}_{0}i}{4\mathrm{\pi }d}$ (Perpendicular to the plane in outward direction)
$\Rightarrow B_{\mathrm{net}}=\frac{{\mathrm{\mu }}_{0}i}{4\mathrm{\pi }d}$ (Perpendicular to the plane in outward direction)

At point Q, the magnetic field due to wire W₁ is given by
$B_1=\frac{{\mathrm{\mu }}_{0}i}{4\mathrm{\pi }d}$ (Perpendicular to the plane in inward direction)
At point Q, the magnetic field due to wire W₂ is given by
B₂ = 0
$\Rightarrow B_{\mathrm{net}}=\frac{{\mathrm{\mu }}_{0}i}{4\mathrm{\pi }d}$ (Perpendicular to the plane in inward direction)

At point R, the magnetic field due to wire W₁ is given by
B₁ = 0
At point R, the magnetic field due to wire W₂ is given by
$B_2=\frac{{\mathrm{\mu }}_{0}i}{4\mathrm{\pi }d}$ (Perpendicular to the plane in inward direction)
$\Rightarrow B_{\mathrm{net}}=\frac{{\mathrm{\mu }}_{0}i}{4\mathrm{\pi }d}$ (Perpendicular to the plane in inward direction)

At point S, the magnetic field due to wire W₁ is given by
$B_1=\frac{{\mathrm{\mu }}_{0}i}{4\mathrm{\pi }d}$ (Perpendicular to the plane in outward direction)
At point S, the magnetic field due to wire W₂ is given by
B₂ = 0
$\Rightarrow B_{\mathrm{net}}=\frac{{\mathrm{\mu }}_{0}i}{4\mathrm{\pi }d}$ (Perpendicular to the plane in outward direction)
Hence, the magnitude of the magnetic field at points P, Q, R and S is $\frac{{\mathrm{\mu }}_{0}i}{4\mathrm{\pi }d}$.