A long solenoid has 500 turns. When a current of $2 ext{ A}$ is passed through it, the resulting magnetic fl

Question

A long solenoid has 500 turns. When a current of $2 	ext{ A}$ is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $4 	imes 10^{-3} 	ext{ Wb}$. The self-inductance of the solenoid is:

Accepted Answer

The self-inductance ($L$) of a coil is defined by the relationship between the total magnetic flux linkage ($N\phi_B$) and the current ($I$) flowing through it.

1.  **Formula:**
    $$ N\phi_B = LI $$
    Where:
    *   $N$ = Total number of turns
    *   $\phi_B$ = Magnetic flux linked with each turn
    *   $I$ = Current
    *   $L$ = Self-inductance

2.  **Given Values:**
    *   $N = 500$
    *   $I = 2 	ext{ A}$
    *   $\phi_B = 4 	imes 10^{-3} 	ext{ Wb}$

3.  **Calculation:**
    Rearranging the formula to solve for $L$:
    $$ L = \frac{N\phi_B}{I} $$
    $$ L = \frac{500 	imes (4 	imes 10^{-3})}{2} $$
    $$ L = \frac{2000 	imes 10^{-3}}{2} $$
    $$ L = \frac{2.0}{2} $$
    $$ L = 1.0 	ext{ H} $$

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