A long solenoid has 500 turns. When a current of 2 ampere is passed through it, the resulting magnetic flux linked with each turn of the solenoid is 4 × 10-3 Wb. The self-inductance of the solenoid is

Question

A long solenoid has 500 turns. When a current of 2 ampere is passed through it, the resulting magnetic flux linked with each turn of the solenoid is 4 × 10-3 Wb. The self-inductance of the solenoid is (A) 1.0 henry (B) 4.0 henry (C) 2.5 henry (D) 2.0 henry

Tardigrade Admin · Accepted Answer

For a long solenoid, B=μ0ni=μ0(N/l).i Flux =μ0(N/l).i.A given flux per turn = 4 x 10-3, i = 2 A ∴ Total flux = 4×1 0-3 L= big(μ0(N/l).NA big)=(4×10-3×500/2)=1 henry

Q. A long solenoid has 500 turns. When a current of 2 ampere is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $4 \times 1 0^{- 3}$ Wb. The self-inductance of the solenoid is

1.0 henry

4.0 henry

2.5 henry

2.0 henry

For a long solenoid, $B = μ_{0} ni = μ_{0} \frac{N}{l} . i$
Flux $= μ_{0} \frac{N}{l} . i . A$
given flux per turn = 4 x $1 0^{- 3}$ , i = 2 A
$∴ T o t a l f l ux = 4 \times 1 0^{- 3}$
$L = (μ_{0} \frac{N}{l} . N A) = \frac{4 \times 1 0 ^{- 3} \times 500}{2} = 1$ henry

Q. A long solenoid has 500 turns. When a current of 2 ampere is passed through it, the resulting magnetic flux linked with each turn of the solenoid is 4×10−3 Wb. The self-inductance of the solenoid is