Question

Q-ᴀ ᴄɪʀᴄᴜʟᴀʀ ᴄᴏɪʟ ᴏғ ᴡɪʀᴇ ᴄᴏɴsɪsᴛɪɴɢ ᴏғ 100 ᴛᴜʀɴs, ᴇᴀᴄʜ ᴏғ ʀᴀᴅɪᴜs 8.0 ᴄᴍ ᴄᴀʀʀɪᴇs ᴀ ᴄᴜʀʀᴇɴᴛ ᴏғ 0.40 ᴀ. ᴡʜᴀᴛ ɪs ᴛʜᴇ ᴍᴀɢɴɪᴛᴜᴅᴇ ᴏғ ᴛʜᴇ ᴍᴀɢɴᴇᴛɪᴄ ғɪᴇʟᴅ ʙ ᴀᴛ ᴛʜᴇ ᴄᴇɴᴛʀᴇ ᴏғ ᴛʜᴇ ᴄᴏɪʟ?​

Answer 1

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3.14

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? Hence, the magnitude of the magnetic field is 3.14 × 10–4 T.

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Answer 2

Answer:

❥Number of turns on the circular coil, n = 100

Radius of each turn, r = 8.0 cm = 0.08m

Current flowing in the coil, I = 0.4A

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

$= 4\pi \times {10}^{ - 7} t \: m \: {a}^{ - 1}$

$|b| = 4\pi \times {10}^{ - 7} \times 2\pi \times 100 \times 0.4 \div 4\pi \times 0.08$

$3.14 \times {10}^{ - 4} t$

Hence, the magnitude of the magnetic field is

$3.14 \times {10}^{ - 4} t$

Explanation:

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