Ans: We are given the following:
Length of the wire, $I=3 \mathrm{~cm}=0.03 \mathrm{~m}$
Current flowing in the wire, $1=10 \mathrm{~A}$.
Magnetic field, $B=0.27 \mathrm{~T}$
Angle between the current and magnetic field, $\theta=90^{\circ}$
Magnetic force exerted on the wire could be given as:
$F=B \| \sin \theta$
Substituting the given values, we get,
$$
\begin{array}{l}
\Rightarrow F=0.27 \times 10 \times 0.03 \sin 90^{\circ} \\
\Rightarrow F=8.1 \times 10^{-2} \mathrm{~N}
\end{array}
$$
Therefore, the magnetic force on the wire is found to be $8.1 \times 10^{-2} \mathrm{~N}$ and the direction of the force can be obtained using Fleming’s left-hand rule.