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Derive an expression for the intensity of magnetic field at any point on the axis of a circular current loop. |
Answer» Solution : Let R be the radius of a CURRENT loop, carrying current I. Let R be a POINT on the AXIS of a conductor. Let dB be magnetic field at P, due to a current element .idl. From the figure, `theta + a= 90^@`, so that `alpha = 90^@ – 0` and `cosalpha = cos (90^@ – 0)` `sintheta = R^2/((R^2+x^2)^(1/2))`...(1) Let `dB_x` be the horizontalcomponent of dB. Applying Biot-Savart.s law , We write `dvecB=(mu_0/(4pi))(I(d vecl xx vecr))/r^3` i.e., `dB=(mu_0/(4pi))(Irdl)/r^3 sin theta.` where `theta.=90^@` i.e., `dB=(mu_0/(4pi))(Irdl)/r^3 =(mu_0/(4pi))(Idl)/r^2` ...(2) Component of dB along the horizontalis `dB_x=dBcosalpha` or `dB_x =dB sin theta` `=(mu_0/(4pi))(Idl)/r^2 sin theta` By using (1) we write `dB_x =(mu_0/(4pi))(Idl)/r^2 R/r =(mu_0/(4pi)) (IdlR)/((R^2+x^2)^(3/2))`...(3) or integrating `B_x=int dB_x = int(mu_0/(4pi)) (IR)/((R^2+x^2)^(3/2))dl` i.e., `B_x =(mu_0/(4pi)) (IR)/((R^2+x^2)^(3/2))dl ` i.e., `B_x =(mu_0/(4pi)) (IR)/((R^2+x^2)^(3/2)) (2piR)` where `intdl =2piR` or `B_x=(mu_0/(4pi))((2piIR^2)/((R^2+x^2)^(3/2)))` tesla and `VECB=B_x hati=(mu_0/(4pi)) ((2piIR^2)/((R^2+x^2)^(3/2)))hati`, `dB_y=0` For a CIRCULAR loop , and at the centre, x=0 `vecB=(mu_0/(4pi))((2piI)/R)hati` For a circular conductorcontainingn turns , `B_x hati=(mu_0/(4pi)) (2pinIR^2)/((R^2+x^2)^(3/2))hati` and at the centre , `B_0hati=(mu_0/(4pi)) ((2pinI)/(R))hati` |
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