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`?

विकल्प

• R`sqrt2`

• R`sqrt15`

• 2R

• 3R

Answer 1

R`sqrt15`

Explanation:

Magnetic field at the centre: 

B_c = `(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"`