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A segment `AB` of wire carrying current `I_1` is placed perpendicular to a long straight wire carrying current `I_2` as shown in figure. The magnitude of force experieced by the straight wire `AB` is A. `(mu_0I_1I_2)/(2pi)ln 3`B. `(mu_0I_1I_2)/(2pi) ln 2`C. `(2mu_0I_1I_2)/(2pi)`D. `(mu_0I_1I_2)/(2pi)` |
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Answer» Correct Answer - B At a distance X from current `I_2` `B=mu_0/(2pi)I_2/X` Magnetic force of small element `dX` of wire `AB` `dF=I_2(dX)Bsin90^@` `:. F=int_(x=a)^(x=2a) dF` |
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