Solution Found!
(Ill) A toroid is a solenoid in the shape of a donut (Fig.
Chapter 15, Problem 52P(choose chapter or problem)
(II) A toroid is a solenoid in the shape of a circle (Fig. 20-62). Use Ampère's law along the circular path, shown dashed in Fig. , to determine that the magnetic field
inside the toroid is \(B=\mu_{0} N I / 2 \pi R\), where
is the total number of turns, and
outside the toroid is \(B=0\)
Is the field inside a toroid uniform like a solenoid's? If not, how does it vary?
FIGURE 20-62 Problem 52.
(a) A toroid.
(b) A section of the toroid showing direction of the current for three loops: \(\odot\) means current toward you, and \(\odot\) means current away from you.
Equation Transcription:
Text Transcription:
B=\mu_{0} N I / 2 \pi R
B=0
\odot
\otimes
Questions & Answers
QUESTION:
(II) A toroid is a solenoid in the shape of a circle (Fig. 20-62). Use Ampère's law along the circular path, shown dashed in Fig. , to determine that the magnetic field
inside the toroid is \(B=\mu_{0} N I / 2 \pi R\), where
is the total number of turns, and
outside the toroid is \(B=0\)
Is the field inside a toroid uniform like a solenoid's? If not, how does it vary?
FIGURE 20-62 Problem 52.
(a) A toroid.
(b) A section of the toroid showing direction of the current for three loops: \(\odot\) means current toward you, and \(\odot\) means current away from you.
Equation Transcription:
Text Transcription:
B=\mu_{0} N I / 2 \pi R
B=0
\odot
\otimes
ANSWER:
Step 1 of 3
a)
Every loop of the wire makes a field along its axis. For path (1) with all the loops together, the magnetic field is the same everywhere along the path and parallel to the path. One side of every turn of the wire is enclosed by path 1 and the enclosed current = N1:
