Question

A toroid having a square cross section, 4.76 cm on a side, and an inner radius of 22.0 cm has 740 turns and carries a current of 1.01 A. (It is made up of a square solenoid instead of round one bent into a doughnut shape.) What is the magnitude of the magnetic field inside the toroid at (a) the inner radius and (b) the outer radius?

Answer 1

Answer:

a. $\beta_{in}=6.79x10^{-4}T$

b. $\beta_{out}=5.586x10^{-4}T$

Explanation:

Given:

$N=740$, $I=1.01A$, $u_o=4\pi*10^{-7} T*m/A$.

The magnetic field can be find using the Gauss law in the equation of field magnetic at the inner radius and the outer radius knowing the inner radius and the outer radius are:

$r_{in}=22.0cm*\frac{1m}{100cm}=0.22m$

$r_{out}=r_{in}+4.76=0.22+0.0476=0.2676m$

a.

Magnetic field inner

$\beta_{in}=\frac{u_o*I*N}{2\pi*r}$

$\beta_{in}=\frac{4\pi*10^{-7}T*m/A*1.01A*740}{2\pi*0.22m}$

$\beta_{in}=6.79x10^{-4}T$

b.

Magnetic field outer

$\beta_{out}=\frac{u_o*I*N}{2\pi*r}$

$\beta_{out}=\frac{4\pi*10^{-7}T*m/A*1.01A*740}{2\pi*0.2676m}$

$\beta_{out}=5.586x10^{-4}T$