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A toroid wound with 60.0 turns/m of wire carries a current of 5.00 A. The torus is iron, which has a magnetic permeability of μm=5000μ0 under the given conditions. H and B inside the iron are
  • a)
    340AT/m,2.88T
  • b)
    340AT/m,1.88T
  • c)
    380AT/m,1.98T
  • d)
    300AT/m,1.88T
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
A toroid wound with 60.0 turns/m of wire carries a current of 5.00 A. ...
B=B0+Bm
= μ0nI+ μmnI
=nI(μ+ μm)
=60x5(5000μ0+ μ0)
B=300(5001μ0)
B=1.88T
H=B/μm
=300x5001μ0/5000μ
=300AT
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Community Answer
A toroid wound with 60.0 turns/m of wire carries a current of 5.00 A. ...
Approximately 4π × 10^-7 H/m.

The magnetic field inside the toroid can be calculated using the formula:

B = μnI

where B is the magnetic field, μ is the magnetic permeability, n is the number of turns per unit length (in this case, 60.0 turns/m), and I is the current.

Plugging in the given values, we get:

B = (4π × 10^-7 H/m) × (60.0 turns/m) × (5.00 A)

B = 4.71 × 10^-3 T

Therefore, the magnetic field inside the toroid is 4.71 × 10^-3 T.
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