Answer:
[tex]42.09\mu N[/tex]
Explanation:
We are given that
Number of wires= N=75
Radius of cylinder=0.420 cm
Current in each wire=1.5 A
We have to find the magnitude of the magnetic force per unit length acting on a wire located at 0.22 cm from the center of the bundle.
r=0.22 cm
Force acting on a wire is given by
Force=F=IBL
[tex]\frac{F}{l}=IB=\frac{\mu_0I^2Nr}{2\pi R^2}[/tex]
Substitute the values in the given formula we get
[tex]\frac{F}{l}=\frac{2\cdot 75\cdot (1.5)^2\cdot 0.220}{(0.42)^2}\times \frac{\mu_0}{4\pi}[/tex]
[tex]\frac{F}{l}=10^{-7}\times 420.92[/tex]
[tex]F/l=42.09 \mu N[/tex]
Hence, force per unit length acting on a wire located 0.22 cm from the center of the bundle=[tex]42.09\mu N[/tex]
