Two long, straight wires are separated by a distance of 32.2 cm. One wire carries a current of 2.75 A, the other carries a current of 4.33 A. (a) Find the force per meter exerted on the 2.75-A wire. (b) Is the force per meter exerted on the 4.33-A wire greater than, less than, or the same as the force per meter exerted on the 2.75-A wire

Respuesta :

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]