A Wire Of Fixed Length Is Wound On A Solenoid Of Length L And Radius R. Its Self Inductance Is Found To Be L. Now, If Same Wire Is Wound On A Solenoid Of Length 1/2 And Radius R/2, Then Self Inductance Will Be

Sol:

$$L = \mu _{0} * n^{2} * V$$

Here:

n is the number of turns per unit length

V → Volume of cylinder

$$\Rightarrow \mu _{0} * \frac{n^{2}}{l^{2}} * \Pi r^{2} * 1$$ $$\Rightarrow L = \frac{\mu _{0}n^{2}\Pi r^{2}}{l}$$

Let length of wire be$$l_{1}$$

$$l_{1} = (2\Pi r) * n \\\Rightarrow n = (\frac{l_{1}}{2\Pi r})\\L = \frac{\mu _{0}\Pi r^{2}}{1} * \frac{l^{2}}{4\Pi^{2} r^{2}} = \frac{\mu _{0} l^{2}}{4\Pi l}\\\Rightarrow L \propto \frac{l}{l} = \frac{L_{2}}{L_{1}} = \frac{l}{L_{2}}=\frac{l}{\frac{l}{2}} = 2$$

Therefore, $$L_{2} = 2L_{1}$$

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