Answer:
a)[tex]\frac{F_1}{L}=1.95*10^-^5N[/tex]
b)[tex]\frac{F_2}{L}=1.95*10^-^5N[/tex]
Explanation:
From the question we are told that:
Distance between wires [tex]d=32.2[/tex]
Wire 1 current [tex]I_1=2.75[/tex]
Wire 2 current [tex]I_2=4.33[/tex]
a)
Generally the equation for Force on [tex]l_1[/tex] due to [tex]I_2[/tex] is mathematically given by
[tex]F_1=I_1B_2L[/tex]
Where
B_2=Magnetic field current by [tex]I_2[/tex]
[tex]B_2=\frac{\mu *i_2}{2\pi d}[/tex]
Therefore
[tex]F_1=I_1B_2L[/tex]
[tex]F_1=I_1(\frac{\mu *i_2*l_1}{2\pi d})L[/tex]
[tex]\frac{F_1}{L} =\frac{4*\pi*10^{-7}*2.75*4.33*100 }{2*\pi*12.2 }[/tex]
[tex]\frac{F_1}{L}=1.95*10^-^5N[/tex]
b)
Generally the equation for Force on [tex]I_2[/tex] due to [tex]I_1[/tex] is mathematically given by
[tex]F_2=I_2B_1L[/tex]
Where
B_1=Magnetic field current by [tex]I_2[/tex]
[tex]B_1=\frac{\mu *I_1}{2\pi d}[/tex]
Therefore
[tex]\frac{F_2}{L} =I_2(\frac{\mu *I_1*I_2}{2\pi d})[/tex]
[tex]\frac{F_2}{L}=1.95*10^-^5N[/tex]