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Elodia [21]
3 years ago
13

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?
Physics
1 answer:
SCORPION-xisa [38]3 years ago
3 0

Answer:

a).β=0.53x10^{-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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