Question

The length of solenoid is I whose windings are made of material of density D and resistivity p. The winding resistance is R. The inductance of solenoid is
[m = mass of winding wire, µ₀ = permeability of free space]

पर्याय

• `mu_0/(4pi"l") ("Rm"/(rho"D"))`

• `mu_0/(2pi"l") ("Rm"/(rho"D"))`

• `mu_0/(2pi"l") ((rho"D")/"Rm")`

• `mu_0/(4pi"l") ((rho"D")/"Rm")`

Answer 1

`mu_0/(4pi"l") ("Rm"/(rho"D"))`

Explanation:

The self-inductance of solenoid,

L = `mu_0"N"^2 "A"/"l"`   ...(i)

where, A is the area of cross-section of solenoid and N Is number of turns.

If xis the length of wire, then

R = `(rho"x")/"A"` and m = AxD

`therefore "Rm" = (rhox)/"A" ("AxD")`

`=> x = sqrt("Rm"/(rho"D"))`   ...(ii)

Also, x = 2πrN = N = `x/(2pi"r") = 1/(2pi"r") sqrt("Rm"/rho"D")`    ...[from Eq. (ii)]

Substituting values in Eq. (i), we get

L = `mu_0(sqrt"Rm")/(2pi"r" sqrt(rho"D"))^2 xx "A"/"l"`

`= mu_0 "Rm"/(4pi^2"r"^2rho"D") * (pi"r"^2)/"l"`

`= mu_0/(4pi"l") ("Rm"/(rho "D"))`