The magnetic force on a current carrying wire placed in uniform magnetic field if the wire is oriented perpendicular to magnetic field, is _________

The magnetic force on a current-carrying wire is given by the formula:

F = I × L × B × sin(θ)

I = current in the wire

L = length of the wire

B = magnetic field strength

θ = angle between wire and magnetic field

If the wire is perpendicular to the magnetic field, θ = 90°.

Since sin(90°) = 1, the force becomes:

F = I × L × B

Considering both sides of the wire (or loop), the effective force doubles:

F = 2 × I × L × B

Hence, the answer is 2ILB.