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lidiya [134]
3 years ago
9

A closely wound search coil has an area of 3.13 cm2, 135 turns, and a resistance of 61.1 Ω. It is connected to a charge-measurin

g instrument whose resistance is 44.4 Ω. When the coil is rotated quickly from a position parallel to a uniform magnetic field to one perpendicular to the field, the instrument indicates a charge of 3.44×10−5 C . What is the magnitude of the magnetic field?
Physics
1 answer:
erastovalidia [21]3 years ago
8 0

Answer:

Explanation:

Let the magnitude of magnetic field be B .

flux passing through the coil's  = area of coil x field x no of turns

Φ = 3.13 x 10⁻⁴ x B x 135 = 422.55 x 10⁻⁴ B .

emf induced = dΦ / dt , Φ is magnetic flux.

current i = dΦ /dt x 1/R

charge through the coil = ∫ i dt

= ∫   dΦ /dt x 1/R dt

= 1 / R ∫ dΦ

= Φ / R

Total resistance R = 61.1 + 44.4 = 105.5 ohm .

3.44 x 10⁻⁵ = 422.55 x 10⁻⁴ B / 105.5

B = 3.44 x 10⁻⁵ x 105.5  / 422.55 x 10⁻⁴

= .86 x 10⁻¹

= .086 T .

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