Respuesta :
Answer:
(a) [tex]64\times 10^{- 4}\ N/m[/tex] and is directed perpendicular to both the magnetic field and force.
(b) Greater
Solution:
As per the question:
No. of wires, N = 100
Radius of the cylinder formed, R = 0.500 cm = 0.005 m
Current in each wire, I = 2 A
Distance from the center, d= 0.200 cm = 0.002 m
Now,
(a) For the magnitude of the magnetic force:
We use Ampere circuital Law, to calculate the magnetic field at a distance d = 0.002 m of an imaginary cylinder:
[tex]\oint B.dR = \mu_{o}I_{enclosed}[/tex]
where
[tex]\mu_{o}[/tex] = permeability in free space
B = Magnetic field
[tex]B\oint dR = \mu_{o}I_{enclosed}[/tex]
[tex]B(2\pi d)= \mu_{o}I_{enclosed}[/tex]
[tex]B= \frac{\mu_{o}I_{enclosed}}{2\pi d}[/tex] (1)
where
[tex]I_{enclosed} = A\times J[/tex]
where
A = Area = [tex]\pi d^{2}[/tex]
J = [tex]\frac{NI}}{\pi R^{2}}[/tex]
Thus
[tex]I_{enclosed} = \pi d^{2}\times \frac{NI}}{\pi R^{2}}[/tex] (2)
Using eqn (1) and (2):
[tex]B= \frac{\mu_{o}NId}{2\pi R^{2}}[/tex]
Now,
The magnetic force is given by:
[tex]\vec{F_{B}} = I(L\times \vec{B})[/tex]
For maximum force:
[tex]F_{B} = ILB[/tex]
Thus force per unit length:
[tex]f =\frac{F_{B}{L} = BI[/tex]
Thus
[tex]f = \frac{\mu_{o}NI^{2}d}{2\pi R^{2}} = \frac{2\times 10^{- 7}\times 100\times 2\times 0.002}{0.005^{2}} = 64\times 10^{- 4}\ N/m[/tex]
(Since, [tex]\frac{\mu_{o}}{4\pi} = 10^{- 7},\ thus\ \frac{mu_{o}}{2\pi} = 2\times 10^{- 7}[/tex])
The force is directed perpendicular to the direction of the magnetic field and force.
(b) We know that the distance over which the force acts is direct proportion with the force:
Force, F ∝ Distance, d
Thus the outer edge of the bundle will experience a greater force than that calculated in part (a)
