Question 221: Electromagnetic Induction

Question

To manufacture a solenoid of length ' $l$ ' and inductance ' $L$ ', the length of the thin wire required is (Diameter of the solenoid is very less than length, $\mu_0=$ permeability of free space)

Accepted Answer

Let us use the formula for the inductance of a long solenoid:
$$ L = \mu_0 \frac{N^2 A}{l} $$
Where
$L$ = Inductance
$\mu_0$ = Permeability of free space
$N$ = Number of turns
$A$ = Cross-sectional area
$l$ = Length of solenoid
Let the radius of the solenoid be $r$ (so $A = \pi r^2$).
Step 1: Express $N$ in terms of known quantities
From the formula:
$$ L = \mu_0 \frac{N^2 \pi r^2}{l} $$
$$ N^2 = \frac{L \, l}{\mu_0 \pi r^2} $$
$$ N = \sqrt{ \frac{L \, l}{\mu_0 \pi r^2} } $$
Step 2: Find total length of wire required
Length of wire per turn = Circumference = $2\pi r$
Total length of wire required, $l_{wire} = N 	imes 2\pi r$
Substituting value of $N$,
$$ l_{wire} = 2\pi r \sqrt{ \frac{L \, l}{\mu_0 \pi r^2} } $$
Step 3: Simplify
Let us take $r$ inside the square root:
$$ l_{wire} = 2\pi \left[ r \sqrt{\frac{L l}{\mu_0 \pi r^2}} ight] $$
$$ l_{wire} = 2\pi \left[\sqrt{ \frac{L l r^2}{\mu_0 \pi r^2} } ight] $$
Since $r^2$ cancels,
$$ l_{wire} = 2\pi \left[\sqrt{ \frac{L l}{\mu_0 \pi} } ight] $$
$$ l_{wire} = 2\sqrt{ \pi } \sqrt{ \frac{L l}{\mu_0} } $$
Or,
$$ l_{wire} = \left[ \frac{4 \pi l L}{\mu_0} ight]^{1/2} $$
Final Answer
:
Option
A
is correct.
$$ \boxed{ \left[ \frac{4 \pi l L}{\mu_0} ight]^{1/2} } $$

Answer

$\left[\frac{4 \pi l L}{\mu_0}\right]^{\frac{1}{2}}$

Answer

$\left[\frac{2 \pi l}{\mu_0 \mathrm{~L}}\right]^{\frac{1}{2}}$

Answer

$\left[\frac{4 \pi \mu_0}{l \mathrm{~L}}\right]^{\frac{1}{2}}$

Answer

$\left[\frac{2 \pi \mu_0 \mathrm{~L}}{l}\right]^{\frac{1}{2}}$

Electromagnetic Induction

To manufacture a solenoid of length ' l ' and inductance ' L ', the length of the thin wire required is (Diameter of the solenoid is very less than length, \mu_0= permeability of free space)

Choose the correct answer: