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

Answer:

a)$\frac{F_1}{L}=1.95*10^-^5N$

b)$\frac{F_2}{L}=1.95*10^-^5N$

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 $I_2$

$B_2=\frac{\mu *i_2}{2\pi d}$

Therefore

$F_1=I_1B_2L$

$F_1=I_1(\frac{\mu *i_2*l_1}{2\pi d})L$

$\frac{F_1}{L} =\frac{4*\pi*10^{-7}*2.75*4.33*100 }{2*\pi*12.2 }$

$\frac{F_1}{L}=1.95*10^-^5N$

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 $I_2$

$B_1=\frac{\mu *I_1}{2\pi d}$

Therefore

$\frac{F_2}{L} =I_2(\frac{\mu *I_1*I_2}{2\pi d})$

$\frac{F_2}{L}=1.95*10^-^5N$