The magnetic field in the solenoid due to current 1.2A is 54.807 x 10⁻⁵ A/m
The No. of Loops in the solenoid, N = 200
Length of the solenoid, l = 55 cm
The current passing through the solenoid, I = 1.2 A
The magnetic field passing through the solenoid can be calculated using the Ampere's law which is
B = μ₀nI
where B is the magnetic field
μ₀ is the permeability of free space
n is the no. of turns per meter
I is the current passing through the solenoid
Thus , we can re-write this formula a
[tex]B = \frac{\mu_0 N I}{l}[/tex]
Let us substitute the know values in the above equation , we get
B = (4 x 3.14 x 10⁻⁷ x 200 x 1.2 ) / (55 x 10⁻²)
where μ₀ = 4π x 10⁻⁷
B = 3014.4 x 10⁻⁷ / 55 x 10⁻²
B = 54.807 x 10⁻⁷ x 10²
B = 54.807 x 10⁻⁵ A/m
Therefore , the magnetic field passing through the given solenoid is
54.807 x 10⁻⁵ A/m
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