Saved Bookmarks
| 1. |
Deduce the relation for the magnetic induction at a point due to an infinitely long straight conductor carrying current. |
|
Answer» <P> Solution :Consider a long straight wire NM with current I flowing from N to M as shown in Fig. Let P be the point at a distance a from point . Consider an element of LENGTH dl of the wire at a distance l from point O and `hat r` be the vector joining the element dl with the pointP. Let `theta` be the angle between `vec(dl)" and " hatr`. Then , the magnetic field at P due to the element is `vec(dB) = (mu_(0)I)/(4pi) (vec(dl))/r^(2) sin theta(" unit vector perpendicular to " vec(dl) and VECR)`The direction of the field is perpendicular to the plane of the paper and going into it. This can be determined by taking the cross product between two vectors `vec(dl)" and " hat r ` ( let it be `hat n`) . The net magnetic field can be determined by integrating equation with PROPER limits . From the Figure, in a right angle triangle PAO,  `TAN ( pi - theta) =a/l` ` l = - a/ (tan theta) ( "since "tan(pi - theta) = - tan theta)` ` l = - a cot theta and r = a cosec theta` Differentiating , `dl = a cosec .^(2) theta d theta` `vec(dB) =( mu_(0)I)/(4 pi) ((a cosec. ^(2) theta d theta))/ ((a cosec theta )^(2)) sin theta hat n ` ` vec(dB) = (mu_(0)I)/(4 pi) ((a cosec . ^(2) theta d theta ))/((a^(2) cosec ^(2) theta)) sin theta hatn ` ` = (mu_(0)I)/(4 pi a) sin theta d theta hat n` This is the magnetic field at a point P due to the current in small elemental length . Note that we have expressed the magnetic field OP in terms of angular coordinate i.e. `theta`. Therefore, the net magnetic field at the point P can be obtained by integrating`vec(dB)` by varying the angle from `theta = varphi _(1) " to " theta = varphi_(2) ` is `vecB = (mu_(0)I)/(4 pi a) underset(varphi_(1))overset(varphi_(2))int sin theta d theta hatn = (mu_(0)I)/(4 pi a) ( cos varphi_(1) - cos varphi_(2)) hat n ` For a an infinitely long straight wire , `varphi_(1) = 0 " and " varphi_(2) = pi ` , the magnetic field is ` vecB = (mu_(0)I)/ (2 pi a) hat n` Note that here `hat n` represents the unit vector from the point O to P . |
|