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Consider two parallel co-axial circular coils of equal radius R, and number of turn 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 (mu_(0)NI)/R, approximately. [Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.] |
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Answer» Solution :Radii of TWO parallel co-axial circular coils = R , Number of turns on each coil = N Current in both coils = l , Distance between both the coils = R Let us consider point Q at distance d from the centre , Then one coil is at a distance of `R/2+d`from point Q `:.` Magnetic field at point Qis given as `B_(1)=(mu_(0)NIR^(2))/(2[(R/2+d)^(2)+R^(2)]^(3//2))` Also, the other coil is at a distance `R/2-d` from point Q `:.` Magnetic field due to this coil is given as: `B_(2)=(mu_(0)NIR^(2))/(2[(R/2-d)^(2)+R^(2)]^(3//2)), `Total magnetic field ,` B=B_(1)+B_(2)` `=(mu_(0)IR^(2))/2[{(R/2-d)^(2)+R^(2)}^(-3//2)]+{(R/2+d)^(2)+R^(2)}^(-3//2)xxN` `=(mu_(0)IR^(2))/2[((5R^(2))/4+d^(2)-Rd)^(-3//2)+((5R^(2))/4+d^(2)+Rd)^(-3//2)]xxN` `=(mu_(0)IR^(2))/2xx((5R^(2))/4)^(-3//2)[(1+4/5 (d^(2))/(R^(2))- 4/5 d/R)^(-3//2)+(1+4/5(d^(2))/(R^(2))+4/5 d/R)^(-3//2)]xxN` For `d lt lt R` neglecting the factor `(d^(2))/(R^(2))` we get `=(mu_(0)IR^(2))/2xx((5R^(2))/4)^(-3//2)xx[(1-(4d)/(5R))^(-3/2)+(1+(4d)/(5R)^(-3//2))]xxN=(mu_(0)IR^(2)N)/(2R^(3))xx(4/5)^(3/2)[1-(6D)/(5R)+1+(6d)/(5R)]` `B=(4/5)^(3//2)(mu_(0)IN)/R=0.72((mu_(0)IN)/R)` Hence, it is proved that the field on the axis around the mid-point between the coils is uniform. |
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