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Question
A straight wire carrying a current of 5 A is bent into a semicircular arc radius 2 cm as shown in the figure. Find the magnitude and direction of the magnetic field at the center of the arc
Solution
The magnetic field at the center of the arc due to straight elements is given by,
`"B" = int mu_°/(4pi) ("i" vec"dl" xx vec"r")/("r"^3)`
For straight elements PQ and RS, `vec"dl" || vec"r"` for the center of the arc.
`∵ vec"dl" xx vec"r" = 0`
⇒ B = 0 .......(Due to PQ & RS)
The magnetic field at the center O, due to the semicircular arc is half of the magnetic field produced by a full circular coil at the center.
`"B" = 1/2 ((mu_°"i")/"2R")`
`"B" =(mu_°"i")/"4R"`.........( Due to semi circular arc)
i = 5 A & R = 2 × 10-2 m
`∴"B" = (4pi xx 10^-7 xx 5)/(4 xx 2 xx 10^-2) = 2.5 pi xx 10^-5 "T"`
⇒ B = 2.5 π × 10-5 T
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