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Question
Derive the expression for magnetic field produced by a current in a circular arc of wire.
Solution
- Consider circular arc of a wire (XY), carrying a current I.
- The circular arc XY subtends an angle θ at the centre O of the circle with radius r of which the arc is a part, as shown in the figure below.
Current carrying wire of a shape of the circular arc. - Consider length element d`vec"l"` lying always perpendicular to `vecr`.
Using Biot-Savart law, the magnetic field produced at O is:
`"d"vec"B" = mu_0/(4pi) ("Id"vec"l" xx vec"r")/"r"^3`
dB = `mu_0/(4pi) "I" ("dlr" sin90^circ)/"r"^3`
= `mu_0/(4pi) ("Idl")/"r"^2` ….(1) - Equation (1) gives the magnitude of the field. The direction of the field is given by the right-hand rule. Thus, the direction of each of the dB is into the plane of the paper. The total field at O is
B = `int"dB" = mu_0/(4pi) "I"int_"A"^θ ("dl")/"r"^2`
= `mu_0/(4pi) "I" int_"A"^θ ("rd"θ)/"r"^2` = `mu_0/(4pi) "I"/"r" θ` ….(2)
where, the angle θ is in radians.
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