JEE Main 12 Jan 2019 Morning Question 8
Question: As shown in, the figure, two infinitely long, identical wires are bent by $ 90{}^\circ $ and placed in such a way that the segments LP and QM are along the x-axis, while segments PS and QN are parallel to the y-axis. If OP=OQ=4cm, and the magnitude of the magnetic field at C) is $ {10^{-4}}T, $ and the two wires carry equal currents (see figure), the magnitude of the current in each wire and the direction of the magnetic field at O will be $ ({\mu_0}=4\pi \times {10^{-7}}N{A^{-2}}) $ [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) 20 A, perpendicular into the page
B) 40 A, perpendicular into the page
C) 20 A, perpendicular out of the page
D) 40 A, perpendicular out of the page.
Show Answer
Answer:
Correct Answer: A
Solution:
The magnetic field due to wires LP and MQ will be zero at point O.
Magnetic field at point O due to the vertical wires is given as
$ B=B _1+B _2=\frac{{\mu_0}}{4\pi }\frac{I}{r}(sin0+sin90)+\frac{{\mu_0}}{4\pi }\frac{I}{r} $ $ (sin0+sin90^{o}) $
$ {10^{-4}}=2\times \frac{4\pi \times {10^{-7}}}{4\pi }\times \frac{I}{0.04}\Rightarrow I=20A $
By right hand thumb rule direction of magnetic field at O will be perpendicular into the page.
