Question

A toroid of 4000 turns has outer radius of 26 cm and inner radius of 25 cm. If the current in the wire is
10 A, calculate the magnetic field of the toroid.​

Answer 1

Answer:

(a)magnetic field outside is zero. It is non zero only inside the core of a toroid.

By definition, Ampere's Law states that integration of magnetic field over an amperian loop is equivalent to constant μ multiplied by current enclosed. Now, the current that flows at outer circumference and inner circumference of toroid are in opposite direction to each other. Therefore, current enclosed by Amperian loop is zero. Thus, magnetic field outside the toroid is zero.

Answer 2

Answer:

The magnetic field of the toroid is 3.14×10⁻²T.

Explanation:

In question the given values are,

The number of turns in the toroid(N)=4000

The outer radius of the toroid(r₁)=26cm

The inner radius of the toroid(r₂)=25cm

The current flowing through the toroid(I)=10A

First, let's calculate the mean radius of the toroid. Which is given as,

$\mathrm{r=\frac{r_1+r_2}{2}}$     (1)

By inserting the required values in equation (1) we get;

$\mathrm{r=\frac{26+25}{2}=25.5cm}$     (2)

And the number of turns per unit length(n) in the toroid is calculated as,

$\mathrm{n=\frac{N}{2\pi r}}$      (3)

When we enter values into equation (3), we get;

$n=\frac{4000}{2\pi \times 25.5\times 10^-^2}$     (4)

And the magnetic field of the toroid is calculated as,

$\mathrm{B=\mu_o nI}$      (5)

μo=permitivity of free space=4π×10⁻⁷H.m⁻¹

By inserting values in equation (5) we get;

$\mathrm{B=4\pi \times 10^-^7\times \frac{4000}{2\pi \times 25.5\times 10^-^2} \times 10}$

$\mathrm{B=3.14\times 10^-^2\ T}$

The magnetic field of the toroid is 3.14×10⁻²T.

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