Saved Bookmarks
| 1. |
Obtain an expression for the magnetic dipole moment of current loop. |
|
Answer» Solution :We know that magnetic induction on the axial line of a circular COIL is `B=(mu_(0) N I R^(2))/(2(R^(2)+X^(2)))^(3//2)` where N = Number of turns in the coil R= Radius of the coil X= Distance from centre of the coil i= Current in a coil If `X gt gt R," Then "B=(mu_(0) Ni R^(2))/(2X^(3))` Multiplying and dividing with `2pi` `B=(mu_(0) Ni R^(2))/(2X^(3))xx(2pi)/(2pi)` `B=(mu_(0))/(4PI). (2Ni A)/(x^(3))rarr(1)` We know that magnetic induction field on the axial line of a bar magnet `B=(mu_(0))/(4pi). (2M)/(x^(3))rarr(2)` Comparing the equations (1) and (2) Magnetic moment (M) = Ni A |
|