A straight wire of length 0.62 m carries a conventional current of 0.7 amperes. What is the magnitude of the magnetic field made by the curr

A straight wire of length 0.62 m carries a conventional current of 0.7 amperes. What is the magnitude of the magnetic field made by the current at a location 2.0 cm from the wire

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  1. Answer:

    Magnetic field at point having a distance of 2 cm from wire is 6.99 x 10⁻⁶ T

    Explanation:

    Magnetic field due to finite straight wire at a point perpendicular to the wire is given by the relation :

    [tex]B=\frac{\mu_{0}I }{2\pi R }\times\frac{L}{\sqrt{L^{2}+R^{2} } }[/tex]      ……(1)

    Here I is current in the wire, L is the length of the wire, R is the distance of the point from the wire and μ₀ is vacuum permeability constant.

    In this problem,

    Current, I = 0.7 A

    Length of wire, L = 0.62 m

    Distance of point from wire, R = 2 cm = 2 x 10⁻² m = 0.02 m

    Vacuum permeability, μ₀ = 4π x 10⁻⁷ H/m

    Substitute these values in equation (1).

    [tex]B=\frac{4\pi\times10^{-7}\times 0.7 }{2\pi \times0.02 }\times\frac{0.62}{\sqrt{(0.62)^{2}+(0.02) ^{2} } }[/tex]

    B = 6.99 x 10⁻⁶ T

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