Answer :
Answer:
[tex]F = 2.55*10^{-4} \ N[/tex]
F = [tex]3.808 * 10^{-6} \ N[/tex]
Explanation:
The magnetic field due to the first wire can be written as:
[tex]B_1 = \frac{\mu _o I_1}{2 \pi (2 d)}[/tex]
[tex]B_1 = \frac{\mu _o I_1}{4 \pi d}[/tex]
The magnetic field due to the second wire is as follows:
[tex]B_2 = \frac{\mu _o I_2}{2 \pi d}[/tex]
The net magnetic field B = [tex]B_1 +B_2[/tex]
[tex]B = \frac{\mu _o I_1}{4 \pi d} +\frac{\mu _o I_2}{2 \pi d} \\ \\ B = \frac{ \mu_o }{2 \pi d}(\frac{I_1}{2}+ I_2)[/tex]
Also; the magnetic force on wire segment l = 2.8 m
[tex]F = I_1lB = \frac{ \mu_o }{2 \pi d}(\frac{I_1}{2}+ I_2)[/tex]
Replacing all the values ; we have :
[tex]F = \frac{ 4 \pi * 10^{-7}*6.8*2.8}{2 \pi * 10*10^{-2}}(\frac{6.8}{2}+ 3.3)[/tex]
[tex]F = 2.55*10^{-4} \ N[/tex]
b)
Magnetic field due to the first wire is :
[tex]B_1 = \frac{\mu _o I_1}{4 \pi d}[/tex]
[tex]B_1 = \frac{4.0*10^{-7}*6.8}{4 \pi *10*10^{-2}}[/tex]
[tex]B_1 = 6.8*10^{-6} \ T[/tex]
Magnetic field due to the second wire is :
[tex]B_2 = \frac{\mu _o I_2}{2 \pi d} \\ \\ B_2 = \frac{4*10^{-7}*3.3}{2 \pi *10*10^{-2}}[/tex]
[tex]B_2 = 6.6*10^{-6}T[/tex]
[tex]B_1>B_2[/tex]
Net magnetic field B = [tex]B_1 - B_2[/tex]
B = [tex](6.8*10^{-6} - 6.6*10^{-6})T[/tex]
B = [tex]2*10^{-7} \ T[/tex]
Magnetic force F = [tex]I_1lB[/tex]
F = [tex]6.6*2.8 *2*10^{-7}[/tex]
F = [tex]3.808 * 10^{-6} \ N[/tex]