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Two solenoids of equal length are each made of 2000 turns of copper wire per meter. Solenoid I has a 5.00 cm radius; solenoid II a 10.0 cm r
Question
Two solenoids of equal length are each made of 2000 turns of copper wire per meter. Solenoid I has a 5.00 cm radius; solenoid II a 10.0 cm radius. When equal currents are present in the two solenoids, the ratio of the magnitude of the magnetic field BIalong the axis of solenoid I to the magnitude of the magnetic field BIIalong the axis of solenoid II, BI/BII, is
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3 years
2021-08-22T07:06:53+00:00
2021-08-22T07:06:53+00:00 1 Answers
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Answer:
BI/BII = 1
Explanation:
The magnetic field due to a solenoid is given by the following formula:
where,
B = Magnetic Field due to solenoid
μ = permeability of free space
n = No. of turns per unit length
I = current passing through the solenoid
Now for the first solenoid:
For the second solenoid:
Dividing both equations:
here, no. of turns and the current passing through each solenoid is same:
n₁ = n₂ and I₁ = I₂
Therefore,
BI/BII = 1