Question

The magnitude of the force vector acting on a unit length of a thin wire carrying a current $I=8 \,A$ at a point $O$, if the wire is bent as shown in the figure with a radius, $R=10 \,\pi\, cm$ is

• A. $64 \,\mu N / m$
• B. $32\, \mu N / m$
• C. $20 \,\mu N / m$
• D. $100 \,\mu N / m$

Answer 1

Answer: (A)

Given, current flowing through wire, $I=8\, A$
and radius of circular wire,
$R=10 \,\pi \,cm =10 \pi \times 10^{-2} m$
According to the question,

$\because$ Magnetic field produced by semi-circular current carrying thin wire at centre $O$.
$B=\frac{\mu_{0} I}{4 R}$
Putting the given values, we get
$\Rightarrow B =\frac{4 \pi \times 10^{-7} \times 8}{4 \times 10 \pi \times 10^{-2}} $
$B =8\times 10^{-6} T$
$\therefore $ Magnitude of the force on thin wire per unit length.
$F=I B$
Putting the given values,
$=8 \times 8 \times 10^{-6} $
$=64 \times 10^{-6} N / m =64\, \mu N / m$