If the current flowing in the current-carrying conductor placed in a magnetic field is increased, the magnitude of force will:

Hint:The current is the rate of change of the charge. When a charge moves with some velocity in a magnetic field, the force applied on the charge acts on the plane perpendicular to the plane of the charge and the magnetic field. The force can be represented in terms of the current. From this equation, the relation between the current and the force can be determined.

Formula used:
The current, $I = \dfrac{{dq}}{{dt}}$
The force acting on the charge \[q\] moving with velocity \[v\] in a magnetic field \[\overrightarrow B \] ,
$\overrightarrow F = q\overrightarrow v \times \overrightarrow B $
The velocity vector, $\overrightarrow v = \dfrac{{d\overrightarrow l }}{{dt}}$

Complete step-by-step solution:
In a magnetic field vector \[\overrightarrow B \], the force applied on a charge \[q\] moving with velocity \[v\]is,
$\overrightarrow F = q\overrightarrow v \times \overrightarrow B $, this force is known as the Lorentz force.
If the current flows in a circuit, then the drift of the free charges along the circuit is the reason for the flowing of the current. Generally, in the magnetic field, the applied magnetic force on the free electric charges acts as an applied force on the conductor.
Let, current in the circuit = $I$
The magnetic field in the small portion of the circuit $d\overrightarrow l $is, \[\overrightarrow B \].
Now if $dq$ the charge covers $d\overrightarrow l $ the time $dt$ ,
The current, $I = \dfrac{{dq}}{{dt}}$
And, The velocity vector, $\overrightarrow v = \dfrac{{d\overrightarrow l }}{{dt}}$
Hence, the force on the small portion $d\overrightarrow l $is,
$d\overrightarrow F = dq\overrightarrow v \times \overrightarrow B $
$ \Rightarrow d\overrightarrow F = dq\dfrac{{d\overrightarrow l }}{{dt}} \times \overrightarrow B $
$ \Rightarrow d\overrightarrow F = \dfrac{{dq}}{{dt}}d\overrightarrow l \times \overrightarrow B $
$ \Rightarrow d\overrightarrow F = Id\overrightarrow l \times \overrightarrow B $
$ \Rightarrow dF = BIdl\sin \theta $, $\theta $ is the angle between $d\overrightarrow l $and \[\overrightarrow B \].
Hence, the force is directly proportional to the current for a constant magnetic field.
So, the force will increase if the current in the conductor increases.

 Note:From the equation $ \Rightarrow dF = BIdl\sin \theta $, we can write the magnetic force acting on the whole circuit,
\[F = \int {dF} = \int {BIdl\sin \theta } \]
When the conducting wire is perpendicular to the magnetic field, i.e. $\theta = {90^ \circ }$ hence $\sin \theta = 1$ , the force will be $F = BIl$ .
And, when the conducting wire is parallel to the magnetic field, i.e. $\theta = {0^ \circ }$ hence $\sin \theta = 0$ , the force will be $F = 0$.