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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
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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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Applications of Biot-Savart's Law > Magnetic Field on the Axis of a Circular Current-Carrying Loop
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