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3 years ago

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A toroid having a square cross section, 5.00 cm on a side, and an inner radius of 15.0 cm has 500 turns and carries a current of

0.800 A. (It is made up of a square solenoid—instead of a round one as in Fig. 29-17—bent into a doughnut shape.) What is the magnetic field inside the toroid at (a) the inner radius and (b) the outer radius?

1 answer:

SCORPION-xisa [38]3 years ago

3 0

Answer:

a).β=0.53 $x10^{-3}$ T

a).β=0.40 $x10^{-4}$ T

Explanation:

The magnetic field at distance 'r' from the center of toroid is given by:

$\beta =\frac{u_{o}*I*N}{2\pi*r}$

a).

$N=500\\I=0.800A\\r=15cm*\frac{1m}{100cm}=0.15m\\u_{o}=4\pi x10^{-7}\frac{T*m}{A} \\\beta=\frac{4\pi x10^{-7}\frac{T*m}{A}*0.8A*500}{2\pi*0.15m} \\\beta=0.53x10^{-3}T$

b).

The distance is the radius add the cross section so:

$r_{1}=15cm+5cm\\r_{1}=20cm$

$r_{1} =20cm*\frac{1m}{100cm}=0.20m$

$\beta =\frac{u_{o}*I*N}{2\pi*r1}$

$\beta =\frac{4\pi x10^{-7}*0.80A*500 }{2\pi*0.20m} \\\beta=0.4x10^{-3} T$

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