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A circular loop of radius R carries a current I. Obtain an expression for the magnetic field at a point on its axis at a distance x from its centre. |
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Answer» Solution :Consider a circular loop of radius R, carrying a current I. Plane of the coil is Y-Z plane (i.e., perpendicular to the plane of paper), while the AXIS of the loop OX LIES in the plane of paper. Consider the circular loop to be divided into large number of current elements `Ivecdl`. Two such elements `N_(1)M_(1) and N_(2)M_(2)` diametrically opposite to each other, produce the magnetic fields `vec(dB)_(1)` and `vec(dB)_(2)` perpendicular to both `vecr and vecdl` and given by right hand RULE as, `|vec(dB)_(1)|=|vec(dB)_(2)|=(mu_(0))/(4pi)*(I dlsin90^(@))/(r^(2))=(mu_(0))/(4pi) (Idl)/((R^(2)+x^(2)))` These fields may be resolved into components along x-axis and y-axis. Obviously components along y-axis balance each other but components along x-axis are all summed up. Hence, magnetic field due to whole current loop will be: `B=oint dB sin phi= oint (mu_(0))/(4pi) (Idl)/((R^(2)+x^(2)))*(R )/(sqrt((R^(2)+x^(2))))= (mu_(0)IR)/(4pi (R^(2)+x^(2))^(3//2)) oint DL` `=(mu_(0)IR)/(4pi (R^(2)+x^(2))^(3//2))* 2pi R= (mu_(0)IR^(2))/(2(R^(2)+x^(2))^(3//2))` 
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