Derive the equation for the force acting on two parallel current … - Vidyadip

Question

Derive the equation for the force acting on two parallel current carrying wires.

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Answer

→As shown in the figure, two long parallel wires $a$ and $b$ separated by a distance $d$ and carrying currents $I _a$ and $I _b$ respectively.

→A uniform magnetic field $\vec{B}_a$ is produced by the conductor ' $a$ ' at every point on the conductor ' $b$ '
→The right-hand rule indicates that the direction of the field is downwards.
→From Ampere's circuital law,
$B _a=\frac{\mu_0 I _a}{2 \pi d}$
→This magnetic field causes a force on wire $b$.
→To calculate this force, a segment is considered on the wire $b$ as shown in the figure. The length of the segment is L .
→The force acting on this segment is $F _{b a}= I _b LB _a$ Putting the value from equation (1)
$\begin{aligned}
F _{b a} & = I _b L\left(\frac{\mu_0 I _a}{2 \pi d}\right) \\
\therefore \quad F _{b a} & =\frac{\mu_0 I _a I _b L}{2 \pi d}
\end{aligned}$
→The direction of this force is from $b$ to $a$
→Similarly, the force exerted by wire $b$ on wire $a$ of a segment of length $L$ is
$F _{a b}=\frac{\mu_0 I _a I _b L}{2 \pi d}$
→The direction of this force is from $a$ to $b$. Thus, $\overrightarrow{ F }_{a b}=-\overrightarrow{ F }_{b a}$
→Both the forces are equal in magnitude and opposite in direction. This force is attractive.
→ So, if current flows in the same direction in both wires, the force is attractive and if current flows in opposite direction, the force is repulsive.

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