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There are two parallel long straight wires of radii r and separation between their axes is kept d. Current loop is formed by joining the wires at ends. Current I flows in both parallel wires in opposite directions. Neglect the end effects and also neglect flux inside the wires. Calculate self-inductance per unit length of this system of wires. |
Answer» Solution : Consider an area segment of length l and width dx at a distance x from one of the wires as shown in FIGURE. Distance of this area segment from the other wire will be d -x because SEPARATION between wires is d. All DISTANCES are from the axes of wires. To calculate magnetic flux LINKED with this area segment we need magnetic field INTENSITY. Magnetic field intensity at the site of area segment can be written as follows: `B=(mu_(0) i)/(2pix)+(mu_(0) i)/(2pi(d-x))` Magnetic fields due to both the wires are added because the two are in same direction. Area of this segment is ldx, and we can multiply this with magnetic field intensity to get magnetic flux linked with this area segment. `dphi=Bldx=[(mu_(0) i)/(2pix)+(mu_(0) i)/(2pi(d-x))](ldx)` `implies dphi=(mu_(0) il)/(2pi)[(1)/(x)+(1)/((d-x))]dx` We can integrate the aboverelation to get total flux in between the two parallel wires. `implies phi=(mu_(0) il)/(2pi)int_(r)^(d-r)[(1)/(x)+(1)/((d-x))]dx` `implies phi=(mu_(0) il)/(2pi)["In x - In(d-x)"]_(r)^(d-r)` `implies phi=(mu_(0) il)/(2pi)[{"In (d-r) - In(d - d + r)"}-{"In r - In(d-r)"}]` `phi=(mu_(0) il)/(2pi)[{"In (d-r) - Inr"}-{"In r - In(d-r)"}]` `implies phi=(mu_(0) il)/(2pi)["In (d-r) - Inr - Inr + In(d-r)"]` `phi=(mu_(0) il)/(2pi)["2In(d-r) - 2Inr"]` `implies phi=(mu_(0) il)/(2pi)"In"((d-r)/(r))` We know that `phi=Li` Hence self-inductance for length l of this system can be written as follows: `implies Li=(mu_(0) il)/(pi)"In"((d-r)/(r))` `implies L=(mu_(0) l)/(pi)"In"((d-r)/(r))` `:.` Self-inductance per unit length can be written as `implies (L)/(l)=(mu_(0))/(pi)"In"((d-r)/(r))` |
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