Question

A long hollow cylindrical thick-walled conductor (inner radius 1.50cm, outer radius 4.00cm) carries a current of 24.0A distributed uniformly across its cross section. A long thin wire runs down the center of the cylinder and carries a current of 32.0A in the opposite direction. What is the magnitude of the magnetic field 3.00cm from the axis?

Answer 1

Answer:

13.478 × 10⁻⁵T

Explanation:

Given that,

magnetic field at r = 3cm

due to the current passing through the centre of the hollow cylinder

$B_1 = \frac{U_0I_1}{2\pi r}$

$B_1 = 2.0 \times 10^-^7 \times \frac{32.0}{3.00 \times 10^-^2} \\\\= 21.33 \times 10^-^5T$

magnetic field at r = 3cm

due to the current passing through the centre of the hollow cylinder

$B_2 = \frac{U_0I_2}{2\pi r}$

I₂ = current through the hollow cylinder within r₁ = 1.50cm to r₂ = 3.00cm

$I_2 = \frac{24}{\pi (4^2 - 1.5^2)} \times \pi (3^2 - 1.5^2)$

$I_2 = \frac{24 \times 6.75 }{13.75} = 11.782A$

so,

$B_2 = \frac{2 \times 10^-^7 \times 11.782}{3\times 10^-^2} \\\\= 7.855 \times 10^-^5$

So , magnetic of netfield is

= B₂ - B₁

= 13.478 × 10⁻⁵T

Answer 2

Answer:

B = 13.47 * 10⁻⁵ T

Explanation:

The magnetic field will be the difference between the magnetic field due to the current passing through the center of the cylinder and the one due to the current carried by the hollow cylindrical thick-walled conductor.

Magnetic field due to the current passing through a conductor can be given by the equation:  

 $B = \frac{\mu_{0}I }{2 \pi R}$

Let the magnetic field due to the current passing through the center of the cylinder be B₁

 $B_{1} = \frac{\mu_{0}I_{1} }{2 \pi R}$

 $I_{1} = 32 A\\\mu_{0} = 4\pi * 10^{-7} \\r = 3 cm = 3 * 10^{-2} m$

 $B_{1} = \frac{4\pi* 10^{-7} * 32 }{2 \pi 3 * 10^{-2} }$

B₁ = 0.0002133 T = 21.33 * 10⁻⁵ T

Let the magnetic field due to the current carried by the hollow cylindrical thick-walled conductor be B₂

 $B_{2} = \frac{\mu_{0}I_{2} }{2 \pi R}$

$I_{2} = \frac{24\pi(a^{2} -r^{2})}{\pi (R^{2}-r^{2} ) }$

$I_{2} = \frac{24\pi(3^{2} - 1.5^{2})}{\pi (4^{2}-1.5^{2} ) }$

$I_{2} = 11.78 A$

$B_{2} = \frac{4\pi*10^{-7} * 11.78 }{2 \pi * 3 * 10^{-2} }$

$B_{2} = 7.86 * 10^{-5} T$

B = B₁ - B₂

B = (21.33 * 10⁻⁵) - (7.86 * 10⁻⁵)

B = 13.47 * 10⁻⁵ T