Saved Bookmarks
| 1. |
A straight wire carrying a current of 5 A is bent into a semicircular arc of radius 2 cm as shown in the fig. Find the magnitude and direction of the magnetic field at the centre of the arc. |
|
Answer» Solution :Here current `I = 5 A`, radius of semi circular arc `R = 2 CM = 0.02 m`. The magnetic field due to long straight wire is zero because the centre point of the arc lies on the extended line of wire itself. And field due to semi circular `arc B = 1/2 [(mu_0 I)/(2R)] = (mu_0 I)/(4 R) = ((4 pi XX 10^(-7)) xx 5)/(4 xx 0.02) = 7.85 xx 10^(-5) T`. As per right hand THUMB rule, the magnetic field `vecB` points PERPENDICULAR to plane directed into the paper. |
|