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A tightly wound toroid of inner radius 1 cm and outer radius 2 cm has 890 turns of wire and carries a current of 2.1 A. What is the magnetic field a distance 1.5 cm from the center?

User Cons
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2 Answers

6 votes

Answer: B = 0.025 T

Explanation: The formulae for the magnetic field strength of a toroid is given below as

B = uNI/2πr

Where B = strength of magnetic field =?

u = permeability of free space = 1.256×10^-6 mkg/A²s²

I = current flowing through the toroid = 2.1A

r = distance at which magnetic field is created = 1.5cm = 0.015m

N = number of turns = 890

By substituting the parameters, we have that

B = 1.256×10^-6 × 890 × 2.1 /2 ×3.142 × 0.015,

B = 0.002347464 / 0.09426

B = 0.025 T

User Artooras
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7 votes

Answer:

B = 0.025T.

Step-by-step explanation:

See attachment below please.

A tightly wound toroid of inner radius 1 cm and outer radius 2 cm has 890 turns of-example-1
User Tevya
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