Saved Bookmarks
| 1. |
A bar magnet 8 cm long is placed in the magnetic meridian with the N pole pointing toward geographical north.Two neutral points separated by a distance of 6 cm are obtained on the equatorial axis of the magnet.If horizontal component of earth's field is 3.2xx10^(-5) T,then pole strength of magnet is |
|
Answer» 5 ab amp cm `r=3cm,B_(H)=3.2xx10^(-5)T,m=?` In case of N to N position neutral point is obtained at equator of dipole.At neutral point `B_(H)=B_(eq)` `B_(H)=(mu_(0))/(4pi)(M)/((r^(2)+l^(2))^(3//2))` `B_(H)=(M)/((r^(2)+l^(2))^(3//2)`(in C.G.S. SYSTEM) `therefore m.2l=(B_(H)(r^(2)+l^(2))^(3//2))/(2l)=(3.2xx10^(-5)xx10^(4)(9+16)^(3//2))/(8)` `m=(3.2xx10^(-1)xx(5^(2))^(3//2))/(8)=5ab A cm` |
|