Find the magnetic field due to current i flowing in an elliptical loop at it’s focus. The equation of ellipse (in polar coordinates as shown) is lr=(1+ecosθ). Here e is eccentricity which is a constant. Take l=50cm,e=0.8,i=2A, if your answer is nπ×10−7T, find n.(In polar co-ordinate system, r refers to distance of point on ellipse from focus F and θ refers to angle formed with major axis as shown in anticlockwise sense. 'l' is length of semi-latus rectum)

dlsinα=rdθB due to dl elementdB=μ04π×Idlsinαr2dB=μ04π×Irdθr2dB=μ04π×Idθr∫dB=μ04π∫IdθrB=μ0I4π∫02π(1+ecosθ)ldθB=8π×10−7T

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Find the magnetic field due to current i flowing in an elliptical loop at it’s focus. The equation of ellipse (in polar coordinates as shown) is $\frac{l}{r} = (1 + e \cos θ) .$ Here $e$ is eccentricity which is a constant. Take $l = 50 c m, e = 0.8, i = 2 A,$ if your answer is $n π \times 10^{- 7} T,$ find $n .$
(In polar co-ordinate system, r refers to distance of point on ellipse from focus F and $θ$ refers to angle formed with major axis as shown in anticlockwise sense. ' $l$ ' is length of semi-latus rectum)

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Detailed Solution

$\begin{array}{l} d l \sin α = r d θ \\ B d u e t o d l e l e m e n t \\ d B = \frac{μ_{0}}{4 π} \times \frac{I d l \sin α}{r^{2}} \\ d B = \frac{μ_{0}}{4 π} \times \frac{I r d θ}{r^{2}} \\ d B = \frac{μ_{0}}{4 π} \times \frac{I d θ}{r} \\ \int d B = \frac{μ_{0}}{4 π} \int \frac{I d θ}{r} \\ B = \frac{μ_{0} I}{4 π} \int_{0}^{2 π} \frac{(1 + e \cos θ)}{l} d θ \\ B = 8 π \times 10^{- 7} T \end{array}$