Question

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

Answer 1

a)
`(F_1)/(L)=1.95*10^-^5N`

b)
`(F_2)/(L)=1.95*10^-^5N`

Step-by-step explanation:

From the question we are told that:

Distance between wires
`d=32.2`

Wire 1 current
`I_1=2.75`

Wire 2 current
`I_2=4.33`

a)

Generally the equation for Force on
`l_1` due to
`I_2` is mathematically given by

`F_1=I_1B_2L`

Where

B_2=Magnetic field current by
`I_2`

`B_2=(\mu *i_2)/(2\pi d)`

Therefore

`F_1=I_1B_2L`

`F_1=I_1((\mu *i_2*l_1)/(2\pi d))L`

`(F_1)/(L) =(4*\pi*10^(-7)*2.75*4.33*100 )/(2*\pi*12.2 )`

`(F_1)/(L)=1.95*10^-^5N`

b)

Generally the equation for Force on
`I_2` due to
`I_1` is mathematically given by

`F_2=I_2B_1L`

Where

B_1=Magnetic field current by
`I_2`

`B_1=(\mu *I_1)/(2\pi d)`

Therefore

`(F_2)/(L) =I_2((\mu *I_1*I_2)/(2\pi d))`

`(F_2)/(L)=1.95*10^-^5N`