Magnetics Ques 77
- Two thin long parallel wires separated by a distance $b$ are carrying a current $i$ ampere each. The magnitude of the force per unit length exerted by one wire on the other is $(1986,2 M)$
(a) $\frac{\mu _0 i^{2}}{b^{2}}$
(b) $\frac{\mu _0 i^{2}}{2 \pi b}$
(c) $\frac{\mu _0 i}{2 \pi b}$
(d) $\frac{\mu _0 i}{2 \pi b^{2}}$
Show Answer
Answer:
Correct Answer: 77.(b)
Solution:
Formula:
Magnetic Force Acting On A Current Carrying Wire:
- Force per unit length between two wires carrying currents $i _1$ and $i _2$ at distance $r$ is given by
$ \frac{F}{l}=\frac{\mu _0}{2 \pi} \frac{i _1 i _2}{r} $
$\text { Here, } \quad i _1 =i _2=i $
$\text { and } \quad r =b $
$ \therefore \quad \frac{F}{l} =\frac{\mu _0 i^{2}}{2 \pi b}$
BETA
