Answer:
The magnitude of the magnetic force on the wire carrying greater current is is 4.85 x 10⁻⁴ N
Explanation:
Given;
distance of separation of the two wires, d = 18 cm
current in the first wire, I₁ = 8 A
current in the second wire, I₂ = 26 A
length of the wire, L = 2.1 m
permittivity of free space, μ₀ = 4π x 10⁻⁷ N·m/A = 1.257 × 10⁻⁶ N·m/A
The magnitude of the magnetic force on the wire carrying greater current is given as;
[tex]F_2 = \frac{\mu_o I_1I_2L_2}{2\pi d } \\\\F_2 = \frac{4\pi *10^{-7} *8*26*2.1}{2\pi *0.18 }\\\\F_2 = 4.85 *10^{-4} \ N[/tex]
Therefore, the magnitude of the magnetic force on the wire carrying greater current is is 4.85 x 10⁻⁴ N
Answer:
F=4.85*10^{-4}N
Explanation:
The magnetic force due to the wire with lower current over the wire with greater current is given by:
[tex]F=ilBsin\theta[/tex] (1)
where i is the current of the wire with greater current, l is the length and B is the magnetic field generated by the other wire. Hence, we have to compute the magnetic field by using the formula:
[tex]B=\frac{\mu_0I}{2\pi r}[/tex]
where I is the current of the wire with lower current (8A), r is the distance between the wires (18cm=18*10^-2m) and mu0 is the space permeability. By replacing we have
[tex]B=\frac{(1.257*10^{-6}Tm/A)(8A)}{2\pi (18*10^{-2}m)}=8.89*10^{-6}T[/tex]
B and l are perpendicular to each other. Hence, by replacing in (1) we obtain:
[tex]F=(26A)(2.1m)(8.89*10^{-6}T)=4.85*10^{-4}N[/tex]
F=4.85*10^{-4}N
hope this helps!!
