Four very long, current carrying wires in the same plane intersect to form a square 40.0 cm on each side, as shown in figure. Find the magnitude and direction of the current I so that the magnetic field at the center of the square is zero.

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

4.0 A toward the bottom of the page.

2.5 A toward the bottom of the page.

3.6 A toward the top of the page.

2.0 A toward the bottom of the page.

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$B \propto I, because distance of centre for all wires is zero. B_{1} = k I_{1} = 10 k \otimes, B_{2} = k I_{2} = 8 k \otimes B_{3} = k I_{3} = 20 k ⊙, B_{4} = k I_{4} Now B_{1} + B_{2} + B_{3} + B_{4} = 0 \Rightarrow (10 k + 8 k - 20 k) \otimes + k I_{4} = 0 \Rightarrow k I_{4} = 2 k \otimes \Rightarrow I_{4} = 2 A towards the bottom.$