Chapter 29: Q49P (page 861)

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 Figure bent into a doughnut shape.) (a) What is the magnetic field inside the toroid at the inner radius and (b) What is the magnetic field inside the toroid at the outer radius?

Short Answer

Expert verified

a) The magnetic field inside the toroid at the inner radiusis $5.33 \times 10^{- 4} T$

b) The magnetic field inside the toroid at the outer radius is $4.00 \times 10^{- 4} T$

Step by step solution

Listing the given quantities

$r_{in} = 15 cm = 0.15 m$

$N = 500$

$\begin{matrix} r_{out} = 15 cm + 5 cm \\ = 20 cm \\ = 0.20 m \end{matrix}$

$i = 0.800 A$

Understanding the concept of magnetic field and toroid

At a point inside the toroid, the magnitude $B$ of the magnetic field is given by equation,

$B = \frac{μ_{0} iN}{2 π} \frac{1}{r}$ (i)

Nis the number of turns and r is the distance of the point from the center of the toroid.

Using that equation, we can calculate the magnetic field inside the toroid at the inner and outer radius.

(a) Calculations of the magnetic field inside the toroid at the inner radius

By using equation (i), we can calculate the magnetic field at the inner radius,

$\begin{matrix} B = \frac{μ_{0} iN}{2 π} \frac{1}{r_{in}} \\ = \frac{1.26 \times 10^{- 6} \times 0.800 \times 500}{6.28} \frac{1}{0.15} \\ = 5.33 \times 10^{- 4} T \end{matrix}$

Thus, the magnetic field inside the toroid at the inner radius is $5.33 \times 10^{- 4} T$ .

(b) Calculations of the magnetic field inside the toroid at the outer radius

Similarly, for outer radius we have,

$\begin{matrix} B = \frac{μ_{0} iN}{2 π} \frac{1}{r_{out}} \\ = \frac{1.26 \times 10^{- 6} \times 0.800 \times 500}{6.28} \frac{1}{0.20} \\ = 4.00 \times 10^{- 4} T \end{matrix}$