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alexdok [17]
3 years ago
14

A tightly wound 1000-turn toroid has an inner radius 1.00 cm and an outer radius 2.00 cm, and carries a current of 1.50 A. The t

oroid is centered at the origin with the centers of the individual turns in the z = 0 plane. Determine the magnitude field at a distance of 1,10 cm from the origin.
Physics
1 answer:
V125BC [204]3 years ago
7 0

Therefore, the magnitude of magnetic field at a distance 1.10cm from the origin is 27.3mT

<u>Explanation:</u>

Given;

Number of turns, N = 1000

Inner radius, r₁ = 1cm

Outer radius, r₂ = 2cm

Current, I = 1.5A

Magnetic field strength, B = ?

The magnetic field inside a tightly wound toroid is given by B = μ₀ NI / 2πr

where,

a < r < b and a and b are the inner and outer radii of the toroid.

The magnetic field of toroid is

B = \frac{u_oNI}{2\pi r}

Substituting the values in the formula:

B (1.10cm) = \frac{(4\pi X 10^-^7 ) ( 1000)(1.5)}{2\pi (1.10) } \\\\

B (1.10cm) = 27.3mT

Therefore, the magnitude of magnetic field at a distance 1.10cm from the origin is 27.3mT

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T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}

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Answer:

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