Two very long, straight and insulated wires are kept at 90∘ angle from each other in xy plane as shown in the figure.
These wires carry current of equal magnitude I, whose directions are shown in the figure. The net magnetic field at point P will be:
(A)2πdμ0I(x^+y^)
(B)πd+μ0I(z^)
(C)Zero✓ Correct
(D)−2πdμ0I(x^+y^)
Answer: (C)
Solution
Each long straight wire produces a field of size B=2πdμ0I at P, which is a perpendicular distance d from each. By the right-hand rule the wire along the x-axis (current in +x^) gives a field at P pointing into the page, −k^, while the wire along the y-axis (current in +y^) gives a field out of the page, +k^. The wires are equidistant from P and carry equal currents, so the two contributions are equal in size and opposite in direction; they cancel, and the net magnetic field at P is zero.
JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Single correct option