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Using Biot Savart's law, derive the expressionfor the magnetic field at a point on theaxis of a circularcurrent loop. |
Answer» Solution : CONSIDER a circular loop of radius R carrying a currentI. Let P be a point at a distancex fromthe centrethe , of the loop on axis.AB is currentelement of lengthdl. The distanceof the pointP from currentelement is r. `therefore` From figure `r=sqrt(R^2+x^2) sin theta=R/r` According to Biot-Savart.s law, magnetic fieldat the PONT P due to the current element AB is `dB =mu_0/(4pi)(Idl)/r^2 "sin"phi` Since `phi=90^@` `dB=mu_0/(4pi)(Idl)/r^2` sin `90^@`( `because ` sin `90^@=1`) `dB=mu_0/(4pi)(I dl)/r^2 ` along PM dB due to the current elementA.B. at the point P is `dB=mu_0/(4pi)(I dl)/r^2` along PN. The magnetic field dB along PM can be resolved into two rectangular components dB cos `phi` along Py and dB sin `phi` along PX. Similarly the magnetic field `d_B` along PN can be resolved into two components . If we consider more smallelementscos `theta`componentscancel each other out . `SigmadB cos phi=0 "" alphaSigmad_B cos theta # 0` `B=intdB sin phi` `B=intmu_0/(4pi)(Idl)/r^2 sin phi` Substituting the values of `r=sqrt(R^2+x^2)`and sin `phi=R/r`in the equation , we get `B=intmu_0/(4pi)(Idl)/(R^2+x^2) R/r = int mu_0/(4pi) (I dl)/(R^2+x^2) R/sqrt(R^2+x^2) =mu_0/(4pi) (IR)/((R^2+x^2)^(3//2)) int dl` But `intdl=2piR to` CIRCUMFERENCEOF the loop `therefore B=(mu_0IR)/(4pi(x^2+R^2)^(3//2))xx2piR` `therefore B=mu_0/(4pi)(2piIR^2)/((x^2+R^2)^(3//2))` |
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