Figure shows three long straight and parallel conductors A, B and C carrying currents 3A, 1A and 2A respectively. A length of 0.5 m of the wire B experiences a force of :

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

5 x 10^–6N from L to R

5 x 10^–6N from R to L

10^–5 from left to right

10^–5 from right to left

answer is C.

(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

$B_{1} = \frac{μ_{0} i_{1}}{2 {πr}_{1}}, B_{2} = \frac{μ_{0} i_{2}}{2 {πr}_{2}}$
i₁=3A, i₂=2A
r₁ = 3×10^–2m, r₂ = 4×10^–2m ,
l = 0.5m; B = (B1–B2) ; F = Bil
$F = \frac{4 π \times 10^{- 7}}{2 π} (\frac{3}{3 \times 10^{- 2}} - \frac{2}{4 \times 10^{- 2}}) \times 0 .5 \times 1$
$= 5 \times 10^{- 6} N$ from left to right

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th