Question

Three coplanar parallel wires, each carrying a current of 10 A along the same direction, are placed with a separation 5.0 cm between the consecutive ones. Find the magnitude of the magnetic force per unit length acting on the wires.

Answer 1

Answer:

5.0 ,cm the megnitude force per unit length

Answer 2

Explanation:

To find:  The magnitude of the magnetic force per unit length acting on the wires.

Step 1:

Let's arrange wires , and as shown in the figure.

Given conditions in the question :

Current Magnitude in each wire, $i_1 = i_2 = i_3 = 10 A$

Due to a parallel current-bearing wire, the magnetic force per unit length is given by

$\frac{F}{l}=\frac{\mu_{o} i_{1} i_{2}}{2 \pi d}$

Step 2:

In case of  $W_1$ wire,

$\frac{F}{l}=\frac{F}{l} b y \text { wire } W_{2}+\frac{F}{l} \text { by wire } W_{3}$  

$=\frac{\mu_{0} \times 10 \times 10}{2 \pi \times 5 \times 10^{-2}}+\frac{\mu_{0} \times 10 \times 10}{2 \pi \times 10 \times 10^{-2}}$

$=\frac{\mu_{0} \times 100}{2 \pi \times 5 \times 10^{-2}}+\frac{\mu_{0} \times 100}{2 \pi \times 10 \times 10^{-2}}$

$=\frac{20 \mu_{0}}{2 \pi \times 10^{-2}}+\frac{10 \mu_{0}}{2 \pi \times 10^{-2}}$

$=\frac{2 \times 10^{-7} \times 20}{10^{-2}}+\frac{2 \times 10^{-7} \times 10}{10^{-2}}$

$=\frac{60 \times 10^{-7}}{10^{-2}}$

$=60 \times 10^{-5}$

$=6 \times 10^{-4}$

Step 3:

In case of  $W_2$ wire,

$\frac{F}{l}=\frac{F}{l} b y \text { wire } W_{1}-\frac{F}{l} b y \text { wire } W_{3}$

$=\frac{\mu_{0} \times 10 \times 10}{2 \pi \times 5 \times 10^{-2}}-\frac{\mu_{0} \times 10 \times 10}{2 \pi \times 10 \times 10^{-2}}$

=0  

Step 4:

In case of  $W_3$ wire,

$\frac{F}{l}=\frac{F}{l} b y \text { wire } W_{1}+\frac{F}{l} \text { by wire } W_{2}$

$=\frac{\mu_{0} \times 10 \times 10}{2 \pi \times 5 \times 10^{-2}}+\frac{\mu_{0} \times 10 \times 10}{2 \pi \times 10 \times 10^{-2}}$

$=\frac{\mu_{0} \times 100}{2 \pi \times 5 \times 10^{-2}}+\frac{\mu_{0} \times 100}{2 \pi \times 10 \times 10^{-2}}$

$=\frac{20 \mu_{0}}{2 \pi \times 10^{-2}}+\frac{10 \mu_{0}}{2 \pi \times 10^{-2}}$

$=\frac{2 \times 10^{-7} \times 20}{10^{-2}}+\frac{2 \times 10^{-7} \times 10}{10^{-2}}$

$=\frac{60 \times 10^{-7}}{10^{-2}}$

$=60 \times 10^{-5}$

$=6 \times 10^{-4}$