$\text{r}=0.02\text{m}$
$\theta=30^\circ$
$\text{i}=2\text{A}$
$\text{B}=4\times10^{-1}\text{T}$
$=500\times1\times3.14\times4\times10^{-4}\times4\times10^{-1}\times\Big(\frac{1}{2}\Big)$
$12.56\times10^{-2}=0.1256\approx0.13\text{N-M}$
14 questions · self-marked practice — reveal the answer and mark yourself.
Force on a semicircular wire
= 2iRB
= 2 × 5 × 0.05 × 0.5
= 0.25N

Let ‘F’ be the force applied due to magnetic field on the wire and ‘x’ be the dist covered.
So, $\text{F}\times\text{l}=\mu\text{mg}\times\text{x}$
$\Rightarrow\text{ibBl}=\mu\text{mgx}$
$\Rightarrow\text{x}\frac{\text{ibBl}}{\mu\text{mg}}$
Radius = l, K.E = K
$\text{L}=\frac{\text{Mv}}{\text{qB}}$
$\Rightarrow\text{l}=\frac{\sqrt{2\text{mk}}}{\text{ql}}$
$\Rightarrow\text{B}=\sqrt{\frac{2\text{mk}}{\text{ql}}}$

Force due to the wire AB and force due to wire CD are equal and opposite to each other.
Thus they cancel each other.
Net force is the force due to the semicircular loop = 2iRB
