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A current circular conducting loop exerts a magnetic field both inside and outside it. The magnetic field at the cetre of a current loop of radius 'R' and carrying a current I is given as : B = (mu_0 I)/(2 R) The magnetic field lines due to a circular current loop form closed loops. The direction of the magnetic field is given by the right hand thumb rule, which states that if one curls the palm of his right hand around the current loop with the fingers pointing in the direction of the current, the right hand thumb will give the direction of the magnetic field. From this law it is clear that the direction of magnetic field vecB is always perpendicular to the direction of flow of current or perpendicular to the plane of circular current loop. Draw the magnetic field lines due to a circular current loop placed in a horizontal plane. An enamelled copper wire of length L is bent in the form of a circular loop and a curretn I is passed through it so that a magnetic field B is get up at its centre. Now the same wire is bent in the form of a circular coil having N turns and same current I is passed through it. What is the magnetic field develop at the centre of this coil? |
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Answer» Solution :Initially radius of current LOOP `R = L/(2 PI)` and magnetic field `B = (mu_0 I)/(2 R) = (2 pi u_0 I)/(2L) = (mu L_0 I)/(L)` FINALLY radius of current loop `R. = L/(2 pi N.)` , are the magnetic field `B. = (mu_0 NI)/(2 R.) = (mu_0 NI.2pi N)/(2L) = (pi u_0 N^2 I)/(L) implies B. = N^2 B` |
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