Test: Force on Current Carrying Conductor - NEET MCQ

As charges and velocities are same
F=q(V×B)
So having the same magnitude of charge and same velocity, they'll experience the same magnitude of force.

Given, i₁​=i₂​=i
∴F=μ_0​i²l​/2πb
Hence, force per unit length is F=μ₀​i²​/2πb

The permeability of free space, μ0, is numerically equal to 4π x 10-7. For air and other non-magnetic materials, the absolute permeability is the same constant. For magnetic materials, absolute permeability is not a fixed constant but varies non-linearly with the flux density.

The long straight wire and side AB  carry current in the same direction, hence will attract each other.
The long straight wire and side CD carry current in the opposite direction, hence will repel each other.
Force on side BC  will be equal and opposite to force on side DA.
Since CD  is farther from the wire than AB,  the force of attraction on  AB  will exceed the force of repulsion on CD.
Hence, there will be a net force of attraction on the loop ABCD and it will move towards the wire.

parallel currents attract from the below figure.
and similarly, in antiparallel currents repel.parallel currents attract from the below figure.
and similarly, in antiparallel currents repel.

Electric dipole at rest can be considered as charge particles at rest. hence, no magnetic field is applied on it.

I₁=3A
I₂=8A
l=5m
r=0.1m
F=?
Force per unit length on the small conductor is given by:
f= μ_o2I₁I₂/4πr
Total force on the length of small conductor is
F=fl
F= μ_o2I₁I₂ l/4πr
F=4πx10^-7x2x3x8x5/4π x 0.1
F=2.4x10^-4N

The direction of force (motion) of a current carrying conductor in a magnetic field is given by Fleming’s left hand rule. 
It states that ‘ If we hold the thumb, fore finger and middle finger of the left hand perpendicular to each other such that the fore finger points in the direction of magnetic field, the middle finger points in the direction of current, then the thumb shows the direction of force (motion) of the conductor.
 

Current in both wires, I = 200 A
Distance between the wires, r = 2 cm = 0.02 m
Length of the two wires, l = 50 cm = 0.5 m
Force between the two wires is given by the relation,
F=μ_oI²/2πr
where, μ_o = permeability of free space = 4π x 10^-7 TmA^-1
So, F = [4π x 10^-7 x (200²)]/[2π x 0.02]
⇒   F   =  0.4 Nm^-1

ma=Bil−mg
a=0.3×100×2×10⁻²−10×6×10⁻³​/ m
a= 0.6−0.06​/6×10⁻³
=100−10
=90ms⁻²