Answer :
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
[tex]B_1 = \frac{U_0I_1}{2\pi r}[/tex]
[tex]B_1 = 2.0 \times 10^-^7 \times \frac{32.0}{3.00 \times 10^-^2} \\\\= 21.33 \times 10^-^5T[/tex]
magnetic field at r = 3cm
due to the current passing through the centre of the hollow cylinder
[tex]B_2 = \frac{U_0I_2}{2\pi r}[/tex]
I₂ = current through the hollow cylinder within r₁ = 1.50cm to r₂ = 3.00cm
[tex]I_2 = \frac{24}{\pi (4^2 - 1.5^2)} \times \pi (3^2 - 1.5^2)[/tex]
[tex]I_2 = \frac{24 \times 6.75 }{13.75} = 11.782A[/tex]
so,
[tex]B_2 = \frac{2 \times 10^-^7 \times 11.782}{3\times 10^-^2} \\\\= 7.855 \times 10^-^5\\[/tex]
So , magnetic of netfield is
= B₂ - B₁
= 13.478 × 10⁻⁵T
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:
[tex]B = \frac{\mu_{0}I }{2 \pi R}[/tex]
Let the magnetic field due to the current passing through the center of the cylinder be B₁
[tex]B_{1} = \frac{\mu_{0}I_{1} }{2 \pi R}[/tex]
[tex]I_{1} = 32 A\\\mu_{0} = 4\pi * 10^{-7} \\r = 3 cm = 3 * 10^{-2} m\\[/tex]
[tex]B_{1} = \frac{4\pi* 10^{-7} * 32 }{2 \pi 3 * 10^{-2} }[/tex]
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
[tex]B_{2} = \frac{\mu_{0}I_{2} }{2 \pi R}[/tex]
[tex]I_{2} = \frac{24\pi(a^{2} -r^{2})}{\pi (R^{2}-r^{2} ) }[/tex]
[tex]I_{2} = \frac{24\pi(3^{2} - 1.5^{2})}{\pi (4^{2}-1.5^{2} ) }[/tex]
[tex]I_{2} = 11.78 A[/tex]
[tex]B_{2} = \frac{4\pi*10^{-7} * 11.78 }{2 \pi * 3 * 10^{-2} }[/tex]
[tex]B_{2} = 7.86 * 10^{-5} T[/tex]
B = B₁ - B₂
B = (21.33 * 10⁻⁵) - (7.86 * 10⁻⁵)
B = 13.47 * 10⁻⁵ T