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CBSE-XI-Physics

35: Magnetic Field due to a Current

with Solutions - page 2
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  • #5
    Consider the situation shown in figure. The straight wire is fixed but the loop can move under magnetic force. The loop will
    (a) remain stationary
    (b) move towards the wire
    (c) move away from the wire
    (d) rotate about the wire.
    Figure
    digAnsr:   b
    Ans : (b) move towards the wire

    Force acting on the wire per unit length carrying current i2 due to the wire carrying current i1 placed at a distance d is given by
    `` F=\frac{{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,o\,}}{i}_{1}{i}_{2}}{2\,\mathrm{\,\pi \,}d}``
    So, forces per unit length acting on sides AB and CD are as follows:
    `` {F}_{\,\mathrm{\,AB\,}}=\frac{{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,o\,}}{i}_{1}{i}_{2}}{2\,\mathrm{\,\pi \,}d}(\,\mathrm{\,Towards\,}\,\mathrm{\,the\,}\,\mathrm{\,wire\,})``
    `` {F}_{\,\mathrm{\,CD\,}}=\frac{{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,o\,}}{i}_{1}{i}_{2}}{2\,\mathrm{\,\pi \,}(d+a)}(\,\mathrm{\,Away\,}\,\mathrm{\,from\,}\,\mathrm{\,the\,}\,\mathrm{\,wire\,})``
    Here, FAB > FCD because force is inversly proportional to the distance from the wire and wire AB is closer to the wire carrying current i1.
    The forces per unit length acting on sides BC and DA will be equal and opposite, as they are equally away from the wire carrying current i1, with current i2 flowing in the opposite direction.
    ∴ FBC = `` -``FDA
    Now,
    Net force:
    `` F={F}_{\,\mathrm{\,AB\,}}\mathit{+}{F}_{\,\mathrm{\,BC\,}}\mathit{+}{F}_{\,\mathrm{\,CD\,}}\mathit{+}{F}_{\,\mathrm{\,DA\,}}``
    `` \Rightarrow F=\frac{{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,o\,}}{i}_{1}{i}_{2}}{2\,\mathrm{\,\pi \,}d}+{F}_{\,\mathrm{\,BC\,}}-\frac{{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,o\,}}{i}_{1}{i}_{2}}{2\,\mathrm{\,\pi \,}(d+a)}-{F}_{\,\mathrm{\,BC\,}}``
    `` \Rightarrow F=\frac{{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,o\,}}{i}_{1}{i}_{2}}{2\,\mathrm{\,\pi \,}}\left(\frac{1}{d}-\frac{1}{d+a}\right)``
    `` \Rightarrow F=\frac{{\,\mathrm{\,\mu \,}}_{\,\mathrm{\,o\,}}{i}_{1}{i}_{2}a}{2\,\mathrm{\,\pi \,}d(d+a)}``
    (Towards the wire)
    Therefore, the loop will move towards the wire.
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