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

Answer 1

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

Hence net magnetic field at the center of the arc will be only due to semicircular arc, and according to right-hand thumb rule, the direction of the magnetic field will downward, inside the paper.