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Apply Biot-Savart law to find the magnetic field due to a circular current carrying loop at a point on the axis of the loop. |
Answer» Solution :1. Consider current carrying loop with radius R as SHOWN in figure.  2. Centre point of loop is O and loop lies on X-axis. 3. We wish to calculate the magnetic field at the point P on this axis. 4. Loop is kept perpendicular to the plane of loop. 5. Let x be the distance of P from the centre o of the loop. 6. The magnitude `dvecB` of the magnetic field due to `dvecl` is given by the Biot-Savart law, `dvecB=mu_(0)/(4pi)(Ivec(dl)xxvecr)/r^(3)` 7. Magnitude of this magnetic field is given by, `|dvecB|=mu_(0)/(4pi)(|Ivec(dl)xxvecr|)/r^(3)` = `mu_(0)/(4pi)(Idlrsintheta.)/r^(3)` where `theta.` = angle between `vec(dl)andvecr" but "vec(dl)_|_vecr` so, `sintheta.="sin"pi/2=1` `therefore|dvecB|=(mu_(0)I)/(4pi)(dl)/r^(2)""...(1)` 8. As shown in diagram `r^(2)=x^(2)+R^(2)` `|dvecB|=mu_(0)/(4pi)(dl)/((x^(2)+R^(2)))""...(2)` 9. Direction of `dvecB` is given by perpendicular to plane of `vec(dl)andvecr`. 10. `dvecB` can be divide in two components, parallel component `dB_(x)=dBsintheta` perpendicular component `dB_(_|_)=dBcostheta` 11. When the components perpendicular to the ring axis are summed over, they cancel out and we obtain null result. Here, `dB_(_|_)=dBcostheta` component due to `vec(dl)` is cancelled by the contribution due to the diametrically opposite `vec(dl)` element shown in figure. Thus, only the parallel component survives. 12. The net contribution along X-direction can be obtained by integrating `dB_(x)=dBsintheta` over the loop, `dB_(x)=dBsintheta""...(3)` From diagram geometry, `sintheta=R/((x^(2)+R^(2))^(1//2))""...(4)` 13. Substitute dB and `costheta` in equation (3) `dB_(x)=mu_(0)/(4pi)(IDL)/((x^(2)+R^(2)))*R/((x^(2)+R^(2))^(1//2))` `thereforedB_(x)=mu_(0)/(4pi)(IdlR)/((x^(2)+R^(2))^(3//2))""...(5)` 14. To integrate this term, we can get total magnetic field in x - direction. `B=intdB_(x)` `B=(mu_(0)IR)/(4pi(x^(2)+R^(2))^(3//2))intdl` `B=(mu_(0)IR)/(4pi(x^(2)+R^(2))^(3//2))(2piR)" "[becauseintdl=2piR]` `thereforeB=(mu_(0)IR^(2))/(2(x^(2)+R^(2))^(3//2))""...(6)` 15. Vector form of magnetic field, `vecB=(mu_(0)IR^(2))/(2(x^(2)+R^(2))^(3//2))hati" "[becausehati" is UNIT vector of x-axis"]` |
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