CBSE Class 12-science Answered
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Asked by jain.pradeep | 06 Dec, 2019, 20:24: PM
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/392379401e55b01fb1e8198f613aacd45deb24b2c67d77.02727219ring.png)
Figure shows the two identical circular loops of radius R, one having linear charge density +λ C/m and
other one having charge density -λ C/m are placed so that centres of loop are ata distance (√3 R ) apart.
Electric field is required at centre point O due to both charged loops.
Let us consider small line elements of length dl as shown in the top figure on both loops.
Their respective charges are +λdl and -λdl . Distance between centre point O and these line elements is
calculated as ( 1.323 R )
As seen from figure , electric field at point due to the line elements +λdl and -λdl gives resultant field field along the direction AO
Figure given at bottom shows even if we consider line elements in opposite direction, net resultant field is along the direction AO.
Magnitude of electric field due to each line element , | E+ | = | E- | = ( λdl )/ [ (4πεo ) (1.323 R)2 ]
Hence Net resultant field due to both the loops = 2 { ( 2 π R λ )/ [ (4πεo ) (1.323 R)2 ] } cos 49
Answered by Thiyagarajan K | 07 Dec, 2019, 09:44: AM
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