Question

# For a circular coil of radius R and N turns carrying current I, themagnitude of the magnetic field at a point on its axis at a distance xfrom its centre is given by,( )202 2 3/2 2IR NBx Rμ=+(a) Show that this reduces to the familiar result for field at thecentre 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 samedirection, and separated by a distance R. Show that the field onthe axis around the mid-point between the coils is uniform overa distance that is small as compared to R, and is given by,0 0.72NIBRμ= , approximately.[Such an arrangement to produce a nearly uniform magneticfield over a small region is known as Helmholtz coils.]

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Solution

## The magnitude of magnetic field at the distance xfrom the centre along the axis is given as B= μ 0 I R 2 N 2 ( x 2 + R 2 ) 3/2 Where, Bis the magnetic field, Nis number of turns, Ris the radius, x is the distance at which the field is to be calculated, I is the current following through the coil and μ 0 is the permeability of free space. By using Biot –Savart Law magnetic field dB due to length dl is calculated. The perpendicular components of the field that is dBcosφ will cancel each other the only remaining component will be along R that is dBsinφ. dB= μ 0 4π Idlsinφ ( R 2 + x 2 ) 1/2 From the ΔPCL, we get sinφ= R R 2 + x 2 Then, dB= μ 0 4π Idl ( R 2 + x 2 ) 1/2 ⋅ R R 2 + x 2 Now, on integrating the above equation over the circle of radius R, we get ∫ l dB = μ 0 4π I ( R 2 + x 2 ) 1/2 R R 2 + x 2 ∫ dl B= μ 0 4π I ( R 2 + x 2 ) 1/2 R R 2 + x 2 ⋅( 2πR ) The magnetic field due to a circular coil having N turns along the axis at the distance xwill be given as, B= μ 0 I R 2 N 2 ( x 2 + R 2 ) 3/2 a) The magnetic field at the centre of the circular coil is given as, B= μ 0 I R 2 N 2 ( x 2 + R 2 ) 3/2 By substituting x=0in the above equation, we get B= μ 0 I R 2 N 2 ( R 2 ) 3/2 = μ 0 IN 2R b) As the direction of the current flowing in the two coils is same so the two coils attract each other and the resultant field will be the sum of the field due to the coil Aand B. B= B A + B B Where, B B = μ 0 I R 2 N 2 [ ( d− R 2 ) 2 + R 2 ] 3/2 B A = μ 0 I R 2 N 2 [ ( d+ R 2 ) 2 + R 2 ] 3/2 Therefore, we can write, B= μ 0 I R 2 N 2 [ ( d+ R 2 ) 2 + R 2 ] 3/2 + μ 0 I R 2 N 2 [ ( d− R 2 ) 2 + R 2 ] 3/2 = μ 0 NI R 2 2 [ 1 ( R 2 4 + d 2 +Rd+ R 2 ) 3/2 + 1 ( R 2 4 + d 2 −Rd+ R 2 ) 3/2 ] As d 2 is very small as compared to the R 2 so we can neglect it. Then, the above equation becomes, B= μ 0 NI R 2 2 [ 1 ( 5 R 2 4 +Rd ) 3/2 + 1 ( 5 R 2 4 −Rd ) 3/2 ] On solving further and taking 5 R 2 4 as common from denominator, we get B= μ 0 NI R 2 2 ( 5 R 2 4 ) 3/2 [ 1 ( 1+ 4 5 d R ) 3/2 + 1 ( 1− 4 5 d R ) 3/2 ] = μ 0 NI R 2 2 ( 5 4 ) 3/2 ⋅ R 3 [ ( 1− 3 2 × 4 5 × d R )+( 1+ 3 2 × 4 5 × d R ) ] = μ 0 NI R ( 5 4 ) 3/2 Thus magnetic field at the mid-point between the coils is uniform over a small distance das compared to Rand it will be, B=0.72 μ 0 NI R

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