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?
A circular coil of wire consisting of 10 | Class 12 Physics Chapter Moving Charges and Magnetism, Moving Charges and Magnetism NCERT Solutions
Q1. 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?
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 × 10–7T 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
Hence, the magnitude of the magnetic field is 3.14 × 10–4T.
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What is the correct answer to: 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??
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...
How do you solve 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? step by step?
Step-by-step explanation:
• Number of turns on the circular coil, n = 100
•
• Radius of each turn, r = 8
• 0 cm = 0
• 08 m
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Question is incomplete because the value of mean radius is not given.but anyway you use the formula to find magnetic field at the center of coil that B= munot NI divided by 2 multiply mean radius
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