asked 132k views
5 votes
A toroid filled with a magnetic substance carries a steady current of 2.91 A. The coil contains 802 turns, has an average radius of 4.33 cm. The magnetic field through the toroid is 0.193137 T. Assume the flux density is constant. What is the magnetic field strength H within the core in the absence of the magnetic substance?

1 Answer

4 votes

Answer:

The magnetic field strength is
H = 8577.2 \ A/m

Step-by-step explanation:

From the question we are told that

The current is
A = 2.91 \ A

The number of turns is
N = 802 \ turns

The radius is
R = 4.33 \ cm = 0.0433 \ m

The magnetic filed is
B = 0.193137 \ T

Generally the magnetic field strength is mathematically represented as


H = ( NI)/( 2 \pi * r )

=>
H = ( 802 * 2.91 )/( 2 * 3.142 * 0.0433 )

=>
H = 8577.2 \ A/m

answered
User KML
by
7.7k points
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