Respuesta :
Answer:
The force unit length of one wire on the other is 4.0 x 10⁻⁵ N/m
Explanation:
Given;
distance between the two wires, r = 5 cm = 0.05 m
current on the first wire, I₁ = 2A
current on the second wire, I₂ = 5A
Force per unit length on two parallel current carrying conductors is given as:
[tex]\frac{F}{L} = \frac{\mu_oI_1I_2}{2\pi r}[/tex]
substitute in the values given into the equation;
[tex]\frac{F}{L} = \frac{4\pi *10^{-7} *2*5}{2\pi *0.05} = 4.0 *10^{-5} \ N/m[/tex]
Therefore, the force unit length of one wire on the other is 4.0 x 10⁻⁵ N/m
Answer:
F/L = 4×10-⁵N/m (force per unit length)
Explanation:
Given I1 = 2.0A, I2 = 5.0A, r = 5.0cm = 5×10-²m
To find: F/L
F/L = μo×I1×I2/2πr
μo =4π×10-⁷T.m/A
r = distance between both wires.
F/L = 4π×10-⁷×2.0×5.0/(2π×5×10-²)
F/L = 4×10-⁵N/m.
This force is attractive as the wires attract each other. Two wires(conductor ) carrying current in the same direction attract each other those that carry current in opposite directions repel each other.