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The magnetic coils of a tokamak fusion reactor are in the shape of a toroid having an inner radius of 0.700 m and an outer radius of 1.30 m. The toroid has 800 turns of large-diameter wire, each of which carries a current of 17.0 kA. (a) Find the magnitude of the magnetic field inside the toroid along the inner radius. T (b) Find the magnitude of the magnetic field inside the toroid along the outer radius. T

2 Answers

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Answer:

(a) 3.886 tesla

(b) 2.09 tesla

Step-by-step explanation:

inner radius, r = 0.7 m

outer radius, R = 1.3 m

current, i = 17 kA = 17000 A

Number of turns, N = 800

(a) The magnetic field is given by


B=(\mu _(0))/(2\pi )* N* (i)/(r)


B=(2 * 10^(-7)* 800 * 17000)/(0.7)

B = 3.886 tesla

(b) The magnetic field is given by


B=(\mu _(0))/(2\pi )* N* (i)/(R)


B=(2 * 10^(-7)* 800 * 17000)/(1.3)

B = 2.09 tesla

User Rajith Delantha
by
6.1k points
3 votes

To solve this problem it is necessary to apply the concepts related to Magnetic field in a toroide.

By definition the magnetic field is defined as


B = (\mu_0 NI)/(2\pi r)

Where,


\mu_0 = Permeability constant in free Space

N = Number of loops

I = Current

r = Radius

PART A) For the internal radio,


B = (4\pi*10^(-7)(800)(17*10^3))/(2\pi 0.7)


B = 3.8857T

PART B) For outside radio,


B = (4\pi*10^(-7)(800)(17*10^3))/(2\pi *1.3)


B = 2.0923T

User Brett Hardin
by
5.3k points