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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Sigridomena 3 years 2021-08-22T07:06:53+00:00 1 Answers 26 views 0

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    2021-08-22T07:08:49+00:00

    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}\\

    BI/BII = 1

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