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CBSE-XI-Physics

35: Magnetic Field due to a Current

with Solutions - page 3
Qstn# iv-6 Prvs-QstnNext-Qstn
  • #6
    A long, straight wire of radius r carries a current i and is placed horizontally in a uniform magnetic field B pointing vertically upward. The current is uniformly distributed over its cross section. (a) At what points will the resultant magnetic field have maximum magnitude? What will be the maximum magnitude? (b) What will be the minimum magnitude of the resultant magnetic field?
    Ans : (a) As the wire in question is carrying current, so it will also generate a magnetic field around it. And for a long straight wire it will be maximum at the mid-point called P.
    Now,
    Magnetic field generated by the current carrying wire`` =\frac{{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,o\,}}i}{2\pi r}``
    Net magnetic field = `` \,\mathrm{\,B\,}+\frac{{\,\mathrm{\,\mu \,}}_{0}i}{2\,\mathrm{\,\pi \,}r}`` (b) Magnetic field B = 0
    when `` r<\frac{{\,\mathrm{\,\mu \,}}_{0}i}{2\,\mathrm{\,\pi B\,}}``
    Clearly,
    B = 0
    when `` r=\frac{{\,\mathrm{\,\mu \,}}_{0}i}{2\,\mathrm{\,\pi B\,}}``
    But when `` r>\frac{{\,\mathrm{\,\mu \,}}_{0}i}{2\,\mathrm{\,\pi B\,}}``,
    Net magnetic field = `` \,\mathrm{\,B\,}-\frac{{\,\mathrm{\,\mu \,}}_{0}i}{2\,\mathrm{\,\pi \,}r}``
    Page No 250:
  • #6-a
    At what points will the resultant magnetic field have maximum magnitude? What will be the maximum magnitude?
    Ans : As the wire in question is carrying current, so it will also generate a magnetic field around it. And for a long straight wire it will be maximum at the mid-point called P.
    Now,
    Magnetic field generated by the current carrying wire`` =\frac{{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,o\,}}i}{2\pi r}``
    Net magnetic field = `` \,\mathrm{\,B\,}+\frac{{\,\mathrm{\,\mu \,}}_{0}i}{2\,\mathrm{\,\pi \,}r}``
  • #6-b
    What will be the minimum magnitude of the resultant magnetic field?
    Ans : Magnetic field B = 0
    when `` r<\frac{{\,\mathrm{\,\mu \,}}_{0}i}{2\,\mathrm{\,\pi B\,}}``
    Clearly,
    B = 0
    when `` r=\frac{{\,\mathrm{\,\mu \,}}_{0}i}{2\,\mathrm{\,\pi B\,}}``
    But when `` r>\frac{{\,\mathrm{\,\mu \,}}_{0}i}{2\,\mathrm{\,\pi B\,}}``,
    Net magnetic field = `` \,\mathrm{\,B\,}-\frac{{\,\mathrm{\,\mu \,}}_{0}i}{2\,\mathrm{\,\pi \,}r}``
    Page No 250: