a straight solenoid 0.5m long has 600 turns. If the magnitude is 2×10^-3 wb/m^2 , find the current in solenoid.
Answers
Answered by
1
Given: A straight solenoid 0.5 m long has 600 turns. The magnitude of magnetic field is 2* 10^-3 Wb/m^2.
To find: Current in the solenoid
Explanation: Length of the solenoid(l)= 0.5 m
Number of turns(N)= 600
Number of turns per unit length(n)
= number of turns/ length
= 600/0.5
= 1200
Permeability constant (u)= 4π * 10^-7
Magnetic field (B)= 2* 10^-3 Wb/m^2
Let current be denoted as i.
The formula used is:
B= uni
=> i = un/B
= 4π*10^-7 * 1200/2*10^-3
= 7540 * 10^-4 A
= 0.7540 A
Therefore, the current in the solenoid is 0.7540 Ampere.
Similar questions