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Derive the expression for magnetic field at a point on the axis of a circular current loop. |
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Answer» Solution :Consider a current loop of radius R carrying a steady current I. let .P. be a point at a distance .Y. from the center of the current loop on its axis. Let .dl. be the current element as shown. The magnetic field due to it is given by BIOT- savart.s law. `dvecB=mu_0/(4r) (I|vec(dl)xxvecr|)/r^3` But `r^2=x^2 +R^2` The displacement vector `vecr` from `vec(dl)` to the AXIAL point P is in the X-Y plane. HENCE `|vec(dl)xxvecr|=rdl`  `therefore dB=mu_0/(4pi) (I|vec(dl)xxvecr|)/r^3 =mu_0/(4pi)(Ir.dl)/r^3 =mu_0/(4pi) (I. dl)/r^2 =mu_0/(4pi) (I dl)/(x^2+R^2)` The component of magnetic field along X-direction is `dB_x = dB cos theta` `cos theta = R/r = R/(x^2+R^2)^(1//2)` `dB=mu_0/(4pi) (I dl)/r^2 R/(x^2+R^2)^(1//2) = mu_0/(4pi) (I dl) /((x^2+R^2)) R/(x^2+R^2)^(1//2)` `therefore dB=(mu_0 IR)/(4pi(x^2+R^2)^(3//2)dl` Therefore the total magnetic field at P due to the loop `B=(mu_0. IRdl)/(4pi(x^2+R^2)^(3//2))` `therefore` As `int dl = 2piR "" B=(mu_0 IR^2)/(2(x^2+R^2)^(3//2))=mu_0/2(IR^2)/((x^2+R^2))^(3//2)` |
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