4.16 For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by,

B = μ 0 I R 2 N 2 ( x 2 + R 2 ) 3 2

(a) Show that this reduces to the familiar result for field at the centre of the coil.

(b) Consider two parallel co-axial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by,

B =0.72 μ 0 N I R , approximately.

[Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]

0 5 Views | Posted 5 months ago
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    Answered by

    Payal Gupta | Contributor-Level 10

    5 months ago

    4.16 Radius of the circular coil = R

    Number of turns on the coil = N

    Current in the coil= I

    Magnetic field at a point on its axis at a distance x is given as:

    B = μ 0 I R 2 N 2 ( x 2 + R 2 ) 3 2

    where μ 0 = Permeability of free space = 4 π × 10 - 7 T m A - 1

    If the magnetic field at the centre of the coil is considered, then x = 0, then

    B = μ 0 I R 2 N 2 ( 0 + R 2 ) 3 2 = μ 0 I N 2 R

    This is the familiar result for magnetic field at the centre of the coil.

    Radius of two parallel co-axial circular coils = R

    Number of turns on each coil = N

    Current in both the coils = I μ 0 I R 2 N 2 { ( R 2 - d ) 2 + R 2 } 3 2 + { ( R 2 + d ) 2 + R 2 } 3 2

    Distance between both the coils = R

    Let us consider point Q at a distance d from the centre.

    Then one coil is at a distance of + d from point Q

    Magnetic field at po

    ...more

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