A circular coil of wire consisting of 100 turns, each of radius 8.0 cm
carries a current of 0.40 A. What is the magnitude of the magnetic
field B at the centre of the coil?
Answers
Answered by
6
Answer:
the magnitude of the magnetic field is 3.14 × 10–4 T.
Explanation:
Number of turns on the circular coil, n = 100
Radius of each turn, r = 8.0 cm = 0.08 m
Current flowing in the coil, I = 0.4 A
Magnitude of the magnetic field at the centre of the coil is given by the relation,
| B | = μ0 2πnI / 4π r
Where,
μ0 = Permeability of free space
= 4π × 10–7 T m A–1
| B | = 4π x 10-7 x 2π x 100 x 0.4 / 4π x 0.08
= 3.14 x 10-4 T
Answered by
0
Answer:nuzero NI/2R= Pi×10-5
Explanation:
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