The magnetic force on a current carrying wire placed in uniform magnetic field if the wire is oriented perpendicular to magnetic field is:
(A) $0$
(B) $ILB$
(C) $2ILB$
(D) $ILB/2$

Hint: The magnetic field due to a flowing charge or current carrying conductor is the product of the magnetic field at that point, the current flowing through the conductor and the length of the wire. It depends on the sine of the angle made by the magnetic field and the length of the wire.

Complete step by step solution:
In a magnetic field a current carrying wire experiences a magnetic force, which is given by –
$\vec F = I(d\vec l \times \vec B)$
Where I is the magnitude of current flowing in the conductor,
$dl$is the length of the wire
And B is the magnetic field.
Since the Force is also a vector quantity, the product between the length of wire and magnetic field is a cross product. Therefore, this formula can be rewritten as-
$\vec F = IBL\operatorname{Sin} \theta $
Where$\theta $is the angle between the magnetic field and the orientation of the wire in that field.
In the question we have,
$\theta = 90^\circ $as the wire is perpendicular to the magnetic field.
Therefore the force is given by,
$\vec F = IBL\operatorname{Sin} 90^\circ $
Since, $\sin 90^\circ = 1$
$F = IBL \times 1$
$F = IBL$

Thus, the answer is option (B).

Additional information
This formula for the force experienced by a conducting wire in a magnetic field is derived from the formula about moving charges experiencing in magnetic field as-
$\vec F = q\left( {\vec v \times \vec B} \right)$
Where q is the value of charge and v is its velocity.
Since the electric current is also a flow of charges, along a definite length, that is why it can be rewritten as-
$q\vec v = dq\dfrac{{dl}}{{dt}} = \dfrac{{dq}}{{dt}}dl = IL$

Note: The direction of the force is given by the right hand rule where if the index finger points In the direction if the current flowing inside the wire, and the middle finger points at the direction of the magnetic field line at the point, the direction of thumb gives the direction of the force experienced by the conductor.