Question

A solid cylindrical conductor is supported by insulating disks on the axis of a conducting tube with outer radius Ra = 7.05cm and inner radius Rb = 4.65cm. The central conductor and the conducting tube carry equal currents of I = 3.95A in opposite directions. The currents are distributed uniformly over the cross sections of each conductor. What is the value of the magnetic field at a distance r = 6.05cm from the axis of the conducting tube?

Answer 1

Answer:

0.00000609180381907 T

Explanation:

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

I = 3.95 A

r = 6.05 cm

Cylinder area

$a=\pi(0.0705^2-0.0465^2)$

Area within r = 6.05 cm

$A=\pi(0.0605^2-0.0465^2)$

Current in the enclosure is given by

$I_1=\dfrac{A}{a}I\\\Rightarrow I_1=\dfrac{\pi(0.0605^2-0.0465^2)}{\pi(0.0705^2-0.0465^2)}\times 3.95\\\Rightarrow I_1=2.10722934473\ A$

Net enclosed current

$I_n=I-I_1\\\Rightarrow I_n=3.95-2.10722934473\\\Rightarrow I_n=1.84277065527\ A$

Magnetic field is given by

$B=\dfrac{\mu_0I_n}{2\pi r}\\\Rightarrow B=\dfrac{4\pi \times 10^{-7}\times 1.84277065527}{2\pi\times 0.0605}\\\Rightarrow B=0.00000609180381907\ T$

The magnetic field strength is 0.00000609180381907 T