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Question
A circular coil carrying current 'I' has radius 'R' and magnetic field at the centre is 'B'. At what distance from the centre along the axis of the same coil, the magnetic field will be `"B"/64`?
Options
R`sqrt2`
R`sqrt15`
2R
3R
MCQ
Solution
R`sqrt15`
Explanation:
Magnetic field at the centre:
Bc = `(µ_0 "nl")/2"R"`
Magnetic field at the axial point:
`"B"_"axis" = (µ_0 "nlR"^2)/(2("R"^2 + "z"^2)^(3/2)`
Given:
`"B"_"axis" = "B"_"c"/64`
`(µ_0 "nlR"^2)/(2("R"^2 + "z"^2)^(3/2) )= (µ_0 "nl")/(64xx 2"R")`
∴ `"R"^2 /(2("R"^2 + "z"^2)^(3/2)) = 1/128"R"`
∴ ` ("R"^2 + "z"^2)^(3/2) = 64"R"^3`
∴ ` ("R"^2 + "z"^2)^(1/2) = 4"R"`
∴ ` "R"^2 + "z"^2 = 16"R"^2`
∴ ` "z"^2 = 15"R"^2`
∴ ` "z" = sqrt15"R"`
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Axial Magnetic Field Produced by Current in a Circular Loop
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