InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
There are two parallel current carrying wires X and Y as shown in figure. Find the magnitude and direction of the magnetic field at points, P, Q and R. |
|
Answer» Here, `I_1=10, I_2=15A, r=5cm=5xx10^-2m` Let `hatn` be the unit vector perpendicular to the plane of paper unwards. The magnetic field at P due to current in wire X is `vecB_1=(mu_0)/(4pi)(2I_1)/(r)hatn=10^-7(2xx10)/(5xx10^-2)hatn=4xx10^-5hatnT` The magnetic field at P due to current in wire Y is `vecB_2=(mu_0)/(4pi)(2I_2)/((r+r+r))(-hatn)=10^-7=(2xx15)/(15xx10^-2)(-hatn)=-2xx10^-5T` The resultant magnetic field at P is `vecB_P=vecB_1+vecB_2=(4xx10^-5)hatn-(2xx10^-5)hatn=2xx10^-5Thatn` `=2xx10^-5T` acting normally outwards. At point Q, both `vecB_1` and `vecB_2` will be acting normally inwards. So `vecB_Q=vecB_1+vecB_2=(mu_0)/(4pi)(2xx10)/(5xx10^-2)(-hatn)+(mu_0)/(4pi)(2xx15)/(5xx10^-2)(-hatn)` `=[10^-7xx(20)/(5xx10^-2)+10^-7xx(30)/(5xx10^-2)](-hatn)=(4xx10^-5+6xx10^-5)(-hatn)` `=10^-4T` acting normally inwards. At point R, `vecB_1` will be acting normally inwards and `vecB_2` will be acting normally outwards. So, `vecB_R=vecB_1+vecB_2=10^-7xx(2xx10)/((5+5+5)xx10^-2)(-hatn)+10^-7xx(2xx20)/((5xx10^-2))(hatn)` `=(-4/3xx10^-5+8xx10^-5)(hatn)=20/3xx10^-5T`, acting normally outwards |
|