Question
A straight wire of length $$0.5\,m$$ and carrying a current of $$1.2\,A$$ is placed in uniform magnetic field of induction $$2T.$$ The magnetic field is perpendicular to the length of the wire. The force on the wire is
A.
$$2.4\,N$$
B.
$$1.2\,N$$
C.
$$3.0\,N$$
D.
$$2.0\,N$$
Answer :
$$1.2\,N$$
Solution :
Force on a current carrying conductor placed in a magnetic field is given by
$$F = i\left( {l \times B} \right) = il\,B\sin \theta $$
where, $$\theta =$$ angle between current elements and magnetic field.
If linear conductor carrying current is placed perpendicular to the direction of magnetic field, $$\left( {\theta = {{90}^ \circ }} \right)$$ it will experience maximum force.
i.e., $${F_{\max }} = ilB$$
Given, $$i = 1.2\,A,\,l = 0.5\,m\,\,{\text{and}}\,B = 2T$$
$$\therefore F = 2 \times 1.2 \times 0.5 = 1.2\,N$$