Force per unit length between two parallel long current carrying wires is =2πdμ0I1I2
Here, force between BC and AB will be same in magnitude.

FBC=FBA=2πdμ0I2
Fres=FBC2+FBA2
⟹Fres=2πd2μ0I2=2πdμ0I2
NEET UG 2017 — Physics Electromagnetism
An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current I along the same direction is shown in Figure. Magnitude of force per unit length on the middle wire B is given by

Held on 30 Apr 2017 · Verified 9 Jul 2026.
2πdμ0I2
πd2μ0I2
πd2μ0I2
2πdμ0I2
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
Two point charges +2μC and -2μC are placed 10 cm apart. The electric field at the midpoint is:
A parallel plate capacitor made of circular plates is being charged such that the surface charge density on its plates is increasing at a constant rate with time. The magnetic field arising due to displacement current is :
Two identical charged conducting spheres $A$ and $B$ have their centres separated by a certain distance. Charge on each sphere is q and the force of repulsion between them is $F$. A third identical uncharged conducting sphere is brought in contact with sphere A first and then with B and finally removed from both. New force of repulsion between spheres $A$ and $B$ (Radii of $A$ and $B$ are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as :
The electric potential at distance r from a point charge q is V = q/(4πε₀r). The electric field E is:
The current passing through the battery in the given circuit, is : 
Work through every NEET UG Electromagnetism PYQ, year by year.