A current I = 20 A is directed along the positive x-axis and perpendicular to a magnetic field. A magnetic force per unit length of 0.16 N/m

Question

A current I = 20 A is directed along the positive x-axis and perpendicular to a magnetic field. A magnetic force per unit length of 0.16 N/m acts on the conductor in the negative y-direction. Calculate the magnitude and direction of the magnetic field in the region through which the current passes. magnitude T direction

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Farah 2 weeks 2021-09-04T10:39:59+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-09-04T10:41:05+00:00

    Answer:

    the magnitude and direction of the magnetic field in the region through which the current passes is 0.008 T and +z direction.

    Explanation:

    given information:

    current, I = 20 A

    magnetic force per unit length, F/L = 0.16 N/m

    the conductor in the negative y-direction

    θ = 90° (perpendicular)

    as we know the formula to calculate magnetic force is

    F = B I L sin θ

    B = F/(I L sin θ)

       = (F/L) (1/I sin θ)

       = 0.16 (1/15 sin 90)

       = 0.008 T

    since F is in the negative y direction, based of the right hand rule the magnetic field is in positive z direction

    0
    2021-09-04T10:41:06+00:00

    Answer:

    Explanation:

    Given:

    current, I = 20 A

    Magnetic force per unit length, F/L

    = 0.16 N/m

    Conductor in the negative y-direction, therefore θ = 90° (perpendicular)

    For a magnetic field,

    F = B I L sin θ

    B = F/(I L sin θ)

    = 0.16 × (1/15 sin 90)

    = 0.008 T

    The field is in the +ve z – direction.

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