Two long parallel wires are separated by 18 cm. One of the wires carries a current of 8 A and the other carries a current of 26 A. The permeabilty of free space is 1.257 × 10−6 N · m/A. Determine the magnitude of the magnetic force on a 2.1 m length of the wire carrying the greater current.

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

a and c

c is 100% but i am confused in a

[tex]For which optical devices does d sometimes have a positive value[/tex]

Ibelieve the correct answer would be c.

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!!