A long solenoid having 1000 turns uniformly distributed over a length of 0.5 m produces a magnetic field of 2.5 ｘ10⁻³ T at the center. Find the current in the solenoid. (Answer: 1 A) 



Given:

Number of turn of Solenoid =  N =  1000 = 1 ｘ 10⁴

Length  = L = 0.5 m

Magnetic Field = B = 2.5 ｘ 10⁻³ T

Permeability of free space = µ₀ = 4 ㄫ ｘ 10⁻⁷ H m⁻¹

To Find:

Current in the Solenoid = I = ?


Solution:

We have the following relation

Using Ampere's Law

B = `\frac {N}{L}` µ₀ I

or

I = `\frac {BL }{Nµ₀}`

by putting the corresponding values

I = `\frac {2.5 ｘ 10⁻³ T ｘ 0.5 m }{1 ｘ 10⁴ ｘ 4 ｘ 3.1416 ｘ 10⁻⁷ H m⁻¹}`

I = `\frac {1.25 ｘ 10⁻³ T m }{12.566 ｘ 10⁻⁴ H m⁻¹}`

I = 0.0995 ｘ 10⁻¹ A

or 

I = 0.995  A ---------------Ans.

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⭐ Henry 'H' is the unit of measurement for Inductance:

1 H = Wb A⁻¹ =  V s A⁻¹ = kg m² s⁻² A⁻²   [Wb= Weber, A=Ampere, V=Volt]


⭐ Weber 'Wb' is the unit of measurement for Magnetic Flux:

1 Wb =  V s = kg m² s⁻² A⁻¹   


⭐ Tesla 'T' is the unit of measurement for Magnetic Field:

1 T =   wb m⁻² = N m⁻¹ A⁻¹ = kg m s⁻² A⁻¹   


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