Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The magnitude of force is [tex]F_R} = 5.33 \ m N[/tex]
Explanation:
From the question we are told that
The current in wire A , B and C are [tex]I _a = I_b =I_c = I= 20 A[/tex]
The distance is [tex]R = 15.0mm = \frac{15}{1000} = 0.015m[/tex]
The length of wire C is [tex]L_c = 200cm = \frac{200}{100} = 2 m[/tex]
Generally the force exerted per unit length that is acting in between two current carrying conductors can be mathematically represented as
[tex]\frac{F}{L} = \frac{\mu_o }{2 \pi} \cdot \frac{I_1 I_2 }{R}[/tex]
Where [tex]\mu_o[/tex] is the permeability of free space with a constant value of
[tex]\mu_o = 4 \pi * 10^{-7} N/A^2[/tex]
When the current in the wire are of the same direction the force is positive and when they are in opposite direction the force is negative
Considering force between A and C
[tex]F_{A -C} = \frac{\mu_o }{2 \pi} \cdot \frac{I_a I_c }{2R} * L[/tex]
Considering force between B and C
[tex]F_{B -C} = \frac{\mu_o }{2 \pi} \cdot \frac{I_b I_c }{R} * L[/tex]
The resultant force is
[tex]F_R = F_{B -C} - F_{A -C} = \frac{\mu_o }{2 \pi} \cdot \frac{I_a I_c }{2R} * L - \frac{\mu_o }{2 \pi} \cdot \frac{I_b I_c }{R} * L[/tex]
[tex]F_R} = \frac{\mu_o }{2 \pi} \cdot \frac{I * I }{R} * L * ({1 - \frac{1}{2} })[/tex]
Substituting values
[tex]F_R} = \frac{4 \pi * 10^{-7}}{2 * 3.142} \cdot \frac{ 20*20 }{0.015} * 2 ({\frac{1}{2} })[/tex]
[tex]F_R} = 5.33 *10^{-3} N[/tex]
[tex]F_R} = 5.33 \ m N[/tex]
