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CBSE-XI-Physics

35: Magnetic Field due to a Current

with Solutions - page 3
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  • #5
    A long, straight wire carrying a current of 1.0 A is placed horizontally in a uniform magnetic field B = 1.0 × 10-5 T pointing vertically upward figure. Find the magnitude of the resultant magnetic field at the points P and Q, both situated at a distance of 2.0 cm from the wire in the same horizontal plane.
    Figure
    Ans : Given:
    Uniform magnetic field, B0 = 1.0 × 10-5 T (Vertically upwards)
    Separation of the point from the wire, d = 2 cm = 0.02 m
    The magnetic field due to the wire is given by
    `` {B}_{\,\mathrm{\,w\,}}=\frac{{\mu }_{0}i}{2\,\mathrm{\,\pi \,}d}=\frac{4\,\mathrm{\,\pi \,}\times {10}^{-7}\times 1}{2\pi \times 0.02}``
    `` \Rightarrow {B}_{\,\mathrm{\,w\,}}=1\times {10}^{-5}\,\mathrm{\,T\,}``

    Now,
    Net magnetic field at point P:
    BP = Bw + B0 = 2 × 10-5 T
    Net magnetic field at point Q:
    BQ = Bw - B0 = 0
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