A circular coil of 200 turns has a radius of 10 cm and carries a current of 2.0 A. (a) Find the magnitude of the magnetic field →B at the centre of the coil. (b) At what distance from the centre along the axis of the coil will the field B drop to half its value at the centre?
(3√4=1·5874...)
Answers
magnitude of magnetic field at the center of the coil is 2.5*10^-3 T
distance on the axis at which magnetic field becomes half of is value at center is 7.7 cm
The problem is solve as in the picture.
where,
μo = permeability of free space,
N = no of turns in the coil
i = current in the coil
R = radius of the coil
r = distance from the center of the coil in the axial direction.
we solve for magnetic field at the center first. using this value, we find the distance r by the general formula
(a) The magnitude of the magnetic field →B at the centre of the coil is
(b) At 7.66cm distance from the centre along the axis of the coil, the field B drop to half its value at the centre.
Explanation:
Given:
Coil turns in loop, n = 200 turns
coil Radius, r = 10 cm
coil Current, i = 2 A
(a) Let the coil's magnetic field at its core be B.
As the magnetic field relationship at the center of a circular coil is given
(b) As magnetic field on the axis of the rotating coil at any point P (say) is given by
Where x is the point distance from coil core.
As the question relates
x = 0.766
The magnetic field will decrease to half its value in the middle if the distance from the middle of the spiral along the spiral axis is equal to 7.66 cm