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Give the expression for magnetic field at a point on the axis of a short magnetic dipole. |
Answer» Solution : I `to` Current in the LOOP `R to ` Radius of the loop X-axis `to` Axis of the loop x `to` Distance OP dl `to` Conducting element of the loop ACCORDING to Biot-Savart.s law, the magnetic field at P is `db=(mu_0I|dlxxr|)/(4pir^3)` `r^2=x^2+R^2` `|dl xx r | =RDL` (`because` they are perpendicular ) `therefore dB=mu_0/(4pi). (IDL)/(x^2+R^2)` i.e., dB has two components `-dB_x` and `dB_1, dB_1` is CANCELLED out and only the x-component remains. `therefore dB_x=dBcos theta` `cos theta =R/(x^2+R^2)^(1//2)` `therefore dBx=(mu_0Idl)/(4pi). d/(x^2+R^2)^(3//2)` Summation of dl over the loop is given by `2piR`. `therefore B=(mu_0 IR^2)/(2(x^2+R^2)^(3//2))hati` |
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