Question

A pair of stationary and infinitely long bent wires are placed in the x-y plane as shown in figure. The wires carry currents of 10A each as shown. The segments L and M are along the x-axis. The segments P and Q are parallel to the y-axis such that OS = OR = 0.02m. Find magnitude and direction of the magnetic induction at the origin O.

• A. $100\,Wb\,m^{-2}$ vertically upward
• B. $10^{-4}\,Wb\,m^{-2}$ vertically downward
• C. $10^{-4}\,Wb\,m^{-2}$ vertically upward
• D. $0.01\,Wb\,m^{-2}$ vertically downward

Answer 1

The Correct Option is C

Point O is along the segments L and M, so magnetic field due to them is zero. Point O is near the end of conductors P and $B_R = B_p + B_Q = \frac{\mu_0 I}{4\pi RO} + \frac{\mu_0 I}{4 \pi SO} = \frac{\mu_0}{4 \pi} \left( \frac{2I}{d}\right)$ Using the given data, $B_R = 10^{-7} \times \frac{2 \times 10}{0.02} = 10^{-4} Wbm^{-2}$ , outside the sheet of paper.

Answer 2

Given:
Two infinitely long wires carrying currents of 10 A each.
They are placed such that segment L (along the x-axis) is 0.02 m away from the origin O, and similarly for segment M.
Segments P and Q are along the y-axis and are symmetrically placed.

We need to find the magnetic field at the origin O.
1. Magnetic Field from Segments L and M (along the x-axis):
Each wire (L and M) contributes a magnetic field perpendicular to the wire and is inversely proportional to the distance from the wire.

2. Direction of the Magnetic Field:
Both wires L and M are parallel and carry currents in opposite directions along the x-axis. According to the right-hand rule for the magnetic field around a wire, the magnetic fields from L and M will point vertically downward at the origin O.

3. Magnitude of the Magnetic Field:
The magnitude B of the magnetic field at a distance r from an infinitely long straight wire carrying current I is given by \(B = \frac{\mu_0 I}{2\pi r}\).

For each wire (L and M):
\(B_{L+M} = \frac{\mu_0 \cdot 10}{2\pi \cdot 0.02} = \frac{10^{-7} \cdot 10}{\pi \cdot 0.02} = \frac{10^{-6}}{\pi \cdot 0.02} = \frac{10^{-6}}{0.0628} = 10^{-5} \text{ T}\)

Therefore, the magnetic field at the origin O due to the wires L and M is \( 10^{-5} \) Tesla vertically downward.
Convert Tesla to Weber per square meter (Wb/m\(^2\)):
\(1 \text{ T} = 10^4 \text{ Wb/m}^2\)
So, \(10^{-5}\)T = \(10^{-1} \times 10^4 \text{ Wb/m}^2 = 10^{-4} \text{ Wb/m}^2\).

So, the correct option is (C): $10^{-4}\,Wb\,m^{-2}$ vertically upward