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Natasha2012 [34]
2 years ago
5

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
Physics
1 answer:
Nookie1986 [14]2 years ago
5 0

Answer:

BI/BII = 1

Explanation:

The magnetic field due to a solenoid is given by the following formula:

B = \mu nI\\

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:

B_1 = \mu n_1I_1 \\

For the second solenoid:

B_2 = \mu n_2I_2\\

Dividing both equations:

\frac{B_1}{B_2} = \frac{\mu n_1I_1}{\mu n_2I_2}\\

here, no. of turns and the current passing through each solenoid is same:

n₁ = n₂ and I₁ = I₂

Therefore,

\frac{B_1}{B_2} = \frac{\mu nI}{\mu nI}\\

<u>BI/BII = 1</u>

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