Saved Bookmarks
| 1. |
A very long wire carrying a current 4sqrt2 A is bent at a right angles. The magnitude of magnetic field (|B|) at a point P lying on a line perpendicular to the bent wire at a distance, d=20 cm from the point of the bending will be (Let mu_(0)=4pixx10^(-7)H//m) |
|
Answer» `1muT` distance, d=20 cm =0.2m and current, `I=4sqrt2A` `:.` MAGNETIC field at a POINT P due to current carrying wire 1,  `B_(1)=(mu_(0))/(pi).(I)/(d)=10^(-7)xx(4sqrt2)/(0.2)` `=2sqrt2xx10^(-6)T` Magnetic field at point P due to current carrying wire 2, `B_(2)=(mu_(0))/(4pi).(I)/(d)=10^(-7)xx(4sqrt2)/(0.2)=2sqrt2xx10^(-6)T` since, `B_(1)andB_(2)` both are in perpendicular direction to each other HENCE the resultant magnetic field at point P, `B=sqrt(B_(1)^(2)+B_(2)^(2))` `=sqrt((2sqrt2xx10^(-6))+(2sqrt2xx10^(-6))^(2))` `=2sqrt2xx10^(-6)sqrt(1+1)` `=4xx10^(-6)T=4muT` Hence, the magnitude of the magnetic field is `4muT.` |
|