V = V1 + V2 + V3 ...(i)
Q = i1V1 = i2V2 = i3V3 ---(ii)
In place of the above arrangement a single capacitor C equ is imagined which can withstand the applied P.D v and can store the same magnitude of charge Q.
Q =CequV ---(iii)
From Eq. (i) + Eq. (iii)
Q = Cequ (V1+V2+V3) ... (iv)
From Eq. (ii) + Eq. (iv)
Adjecent figure shows a point P near a long conductor XY carrying a current I. MN is a short current carrying conductor, kept at the point P, parallel to the conductor XY.
(i) What is the direction of magnetic flux density ‘B’ at the point P due to the current flowing through XY?
(ii) What is the direction of the force experienced by the conductor MN due to the current flowing through XY?