A long solenoid has 1000 turns. When a current of $4 ext{ A}$ flows through it, the magnetic flux linked wit

Question

A long solenoid has 1000 turns. When a current of $4 	ext{ A}$ flows through it, the 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: $N\Phi_B = LI$ .

**Given:**
*   Number of turns ($N$) = 1000
*   Current ($I$) = $4 	ext{ A}$
*   Magnetic flux per turn ($\Phi_B$) = $4 	imes 10^{-3} 	ext{ Wb}$

**Calculation:**
1.  **Calculate Total Flux Linkage:**
    $$ 	ext{Total Flux} = N 	imes \Phi_B $$
    $$ 	ext{Total Flux} = 1000 	imes (4 	imes 10^{-3}) 	ext{ Wb} = 4 	ext{ Wb} $$

2.  **Calculate Self-Inductance ($L$):**
    Using the formula $LI = N\Phi_B$:
    $$ L = \frac{N\Phi_B}{I} $$
    $$ L = \frac{4 	ext{ Wb}}{4 	ext{ A}} $$
    $$ L = 1 	ext{ H} $$

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