Question

An ideal toroidal solenoid (Figure 1) has inner radius r1 = 15.1 cm and outer radius r2 = 18.3 cm. The solenoid has 180 turns and carries a current of 8.40 A.
a) What is the magnitude of the magnetic field at 12.6 cm from the center of the torus?
b) What is the magnitude of the magnetic field at 16.3 cm from the center of the torus?
c) What is the magnitude of the magnetic field at 20.7 cm from the center of the torus?

Answer 1

The solution for the three questions  is mathematically given as

Parts A and C are both zeros.

Part B B = 0.001855Tesla

What is the magnitude of the magnetic field at 12.6 cm from the center of the torus?

Parts A and C are both zeros.

For component A, the magnetic field is zero since 12.6 cm is still inside the toroidal solenoid. Part C has no magnetic field since it is 20.7 cm outside of the toroidal solenoid.

Generally, the equation for  magnetic field is  mathematically given as

B = (mu_0*N*I)/(2*pi*r)

Therefore

B = ((4*pi*10^{-7})*180*8.40)/(2*pi*0.163)

B = 0.001855Tesla

In conclusion, the magnitude of the magnetic field at 16.3 cm from the center of the torus is

B = 0.001855Tesla

Read more about the magnetic field

brainly.com/question/23096032

#SPJ1

Answer 2

Answer:

Part A & C: 0 T

Part B: 1.855*10^-3 T

Explanation:

The formula that models the magnetic field of a toroid is B=μ0*N*I/2π*r .

Note: Toroids keep their magnetic field rotating within the coils that carry current.

Part A & C: Thus the B field magnitude 12.6 cm & 20.7 cm away from the center is 0 T.

Part B: $B=\frac{(4\pi *10x^{-7} )(180)(8.4 A)}{(2\pi )(16.3*10x^{-2} m)}$

B=1.855*10^-3 T