Question

❤️ ϙᴜᴇsᴛɪᴏɴ

ᴀ ᴍᴀɢɴᴇᴛɪᴄ ғɪᴇʟᴅ ɪɴ ᴀ ᴄᴇʀᴛᴀɪɴ ʀᴇɢɪᴏɴ ɪs ɢɪᴠᴇɴ ʙʏ ʙ = ʙᴏ ᴄᴏs(ωᴛ) ^ᴋ ᴀɴᴅ ᴀ ᴄᴏɪʟ ᴏғ ʀᴀᴅɪᴜs ᴀ ᴡɪᴛʜ ʀᴇsɪsᴛᴀɴᴄᴇ ʀ ɪs ᴘʟᴀᴄᴇᴅ ɪɴ ᴛʜᴇ x-ʏ ᴘʟᴀɴᴇ ᴡɪᴛʜ ɪᴛs ᴄᴇɴᴛʀᴇ ᴀᴛ ᴛʜᴇ ᴏʀɪɢɪɴ ɪɴ ᴛʜᴇ ᴍᴀɢɴᴇᴛɪᴄ ғɪᴇʟᴅ (sᴇᴇ ғɪɢ 6.6) . ғɪɴᴅ ᴛʜᴇ ᴍᴀɢɴɪᴛᴜᴅᴇ ᴀɴᴅ ᴛʜᴇ ᴅɪʀᴇᴄᴛɪᴏɴ ᴏғ ᴛʜᴇ ᴄᴜʀʀᴇɴᴛ ᴀᴛ (ᴀ, 0, 0) ᴀᴛ ᴛ = π/2ω, ᴛ = π/ω ᴀɴᴅ ᴛ = 3π/2ω.​

Answer 1

B=Bo cos (ωt ) (along + z axis )

we know that,

flux = BA cos theta

flux = BA cos 0

flux = B=Bo cos (ωt ) (πa²)

E = - d flux/ dt

E = -πa² d[Bo cos (ωt )] /dt

E= π a² Bo ω sin (ωt )

(1)

I = E/ R = π a² Bo ω sin (ωt )/R

I = π a² Bo ω sin (ω π/2ω ) /R

I = π a² Bo ω sin π/2 / R

sin π/ 2 = 1

I = π a² Bo ω /R (along positive y axis)

(2)

I = E/ R = π a² Bo ω sin (ωt )/R

I = π a² Bo ω sin (ω π/ω ) /R

I = π a² Bo ω sin π/ R

sin π = 0

I = 0

(3)

I = π a² Bo ω sin (ω 3π/2ω ) /R

I = π a² Bo ω sin 3 π/2 / R

sin 3π/ 2 = - 1

I = - π a² Bo ω /R ( along negative y axis)

Answer 2

Answer:

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Question:-

➠ᴀ ᴍᴀɢɴᴇᴛɪᴄ ғɪᴇʟᴅ ɪɴ ᴀ ᴄᴇʀᴛᴀɪɴ ʀᴇɢɪᴏɴ ɪs ɢɪᴠᴇɴ ʙʏ ʙ = ʙᴏ ᴄᴏs(ωᴛ) ^ᴋ ᴀɴᴅ ᴀ ᴄᴏɪʟ ᴏғ ʀᴀᴅɪᴜs ᴀ ᴡɪᴛʜ ʀᴇsɪsᴛᴀɴᴄᴇ ʀ ɪs ᴘʟᴀᴄᴇᴅ ɪɴ ᴛʜᴇ x-ʏ ᴘʟᴀɴᴇ ᴡɪᴛʜ ɪᴛs ᴄᴇɴᴛʀᴇ ᴀᴛ ᴛʜᴇ ᴏʀɪɢɪɴ ɪɴ ᴛʜᴇ ᴍᴀɢɴᴇᴛɪᴄ ғɪᴇʟᴅ (sᴇᴇ ғɪɢ 6.6) . ғɪɴᴅ ᴛʜᴇ ᴍᴀɢɴɪᴛᴜᴅᴇ ᴀɴᴅ ᴛʜᴇ ᴅɪʀᴇᴄᴛɪᴏɴ ᴏғ ᴛʜᴇ ᴄᴜʀʀᴇɴᴛ ᴀᴛ (ᴀ, 0, 0) ᴀᴛ ᴛ = π/2ω, ᴛ = π/ω ᴀɴᴅ ᴛ = 3π/2ω.

Answer:−

➠27 ᴀ ᴍᴀɢɴᴇᴛɪᴄ ғɪᴇʟᴅ ʙ ɪs ᴄᴏɴғɪɴᴇᴅ ᴛᴏ ᴀ ʀᴇɢɪᴏɴ ʀ ≤ ᴀ ᴀɴᴅ ᴘᴏɪɴᴛs ᴏᴜᴛ ᴏғ ᴛʜᴇ ᴘᴀᴘᴇʀ (ᴛʜᴇ ᴢ-ᴀxɪs), ʀ = 0 ʙᴇɪɴɢ ᴛʜᴇ ᴄᴇɴᴛʀᴇ ᴏғ ᴛʜᴇ ᴄɪʀᴄᴜʟᴀʀ ʀᴇɢɪᴏɴ. ᴀ ᴄʜᴀʀɢᴇᴅ ʀɪɴɢ (ᴄʜᴀʀɢᴇ = ϙ) ᴏғ ʀᴀᴅɪᴜs ʙ, ʙ > ᴀ ᴀɴᴅ ᴍᴀss ᴍ ʟɪᴇs ɪɴ ᴛʜᴇ x-ʏ ᴘʟᴀɴᴇ ᴡɪᴛʜ ɪᴛs ᴄᴇɴᴛʀᴇ ᴀᴛ ᴛʜᴇ ᴏʀɪɢɪɴ. ᴛʜᴇ ʀɪɴɢ ɪs ғʀᴇᴇ ᴛᴏ ʀᴏᴛᴀᴛᴇ ᴀɴᴅ ɪs ᴀᴛ ʀᴇsᴛ. ᴛʜᴇ ᴍᴀɢɴᴇᴛɪᴄ ғɪᴇʟᴅ ɪs ʙʀᴏᴜɢʜᴛ ᴛᴏ ᴢᴇʀᴏ ɪɴ ᴛɪᴍᴇ Δᴛ. ғɪɴᴅ ᴛʜᴇ ᴀɴɢᴜʟᴀʀ ᴠᴇʟᴏᴄɪᴛʏ ω ᴏғ ᴛʜᴇ ʀɪɴɢ ᴀғᴛᴇʀ ᴛʜᴇ ғɪᴇʟᴅ ᴠᴀɴɪsʜᴇs

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hope it helps you bhai(◍•ᴗ•◍)✌