Saved Bookmarks
| 1. |
Explain briefly how bar magnets act as equivalent solenoids. |
Answer» Solution : The magnitude of magnetic field at a point 'P' due TOA circular element of thickness 'dx' is given by `dB=((mu_(o))/(4pi))(2pindxIa^(2))/(((r-X)^(2)+a^(2))^(1//2))` The magnitude of total magnetic field is given by `B=((mu_(o))/(4pi))2pinIa^(2)int_(-l)^(l)(dx)/([(r-x)^(2)+a^(2)]^(3//2))` For `r gt gt a ` and r `gt gtl, [(r-x)^(2)+a^(2)]^(3//2)=r^(3)` `:. "" B = ((mu_(o))/(4pi))(2pinIa^(2))/(r^(3))int_(-l)^(l)dx "" int_(-l)^(l)dx=(x)_(-l)^(l)` i.e., `"" B = ((mu_(o))/(4pi))(2pinIa^(2))/(r^(3))2L "" = l +l = 2l` We note that the product n `(2l) I (pia^(2))` = Magnetic moment 'm' `:. "" B = ((mu_(o))/(4pi)) (2m)/(r^(3)) "" --(1)` This expression is similar to the magnetic field at a point on the axis of a short magnet. Therefore a bar magnet and a solenoid produce similar magnetic FIELDS. |
|