A 580-turn solenoid is 18 cm long. The current in it is 36 A. A straight wire cuts through the center of the solenoid, along a 2.0-cm diamet

Question

A 580-turn solenoid is 18 cm long. The current in it is 36 A. A straight wire cuts through the center of the solenoid, along a 2.0-cm diameter. This wire carries a 27-A current downward (and is connected by other wires that don’t concern us). What is the magnitude of the force on this wire assuming the solenoid’s field points due east?

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Thu Thảo 5 months 2021-08-25T11:28:33+00:00 2 Answers 2 views 0

Answers ( )

    0
    2021-08-25T11:29:40+00:00

    Answer: The magnitude of the force is 0.079N

    Explanation: Please see the attachments below

    0
    2021-08-25T11:30:03+00:00

    Answer:

    F = 0.078N

    Explanation:

    In order to calculate the magnitude of the force on the wire you first calculate the magnitude of the magnetic field generated by the solenoid, by using the following formula:

    B=\frac{\mu_oNi}{L}         (1)

    μo: magnetic permeability of vacuum = 4π*10^-7 T/A

    N: turns of the solenoid = 580

    i: current in the solenoid = 36A

    L: length of the solenoid = 18cm = 0.18m

    You replace the values of all parameters in the equation (1):

    B=\frac{(4\pi*10^{-7}T/A)(580)(36A)}{0.18m}=0.145T

    Next, you calculate the force exerted on the wire, by using the following formula:

    F=iLBsin\theta         (2)

    i: current in the wire = 27A

    L: length of the wire that perceives the magnetic field (the same as the radius of the solenoid) = 2.0 cm = 0.02m

    θ: angle between wire and the direction of B

    B: magneitc field in the solenoid = 0.145T

    The direction of the wire are perpendicular to the direction of the magnetic field, hence, the angle is 90°.

    You replace the values of the parameters in the equation (2):

    F=(27A)(0.02m)(0.145T)sin90\°=0.078N

    The magnitude of the force on the wire is 0.078N

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