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 = NL µ₀ I
or
I = BLNµ₀
by putting the corresponding values
I = 2.5x10⁻³Tx0.5m1x10⁴x4x3.1416x10⁻⁷Hm⁻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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