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A long straight wire carrying current of `25A` rests on a table as shown in figure. Another wire PQ of length 1m, mass `2*5g` carries the same current but in the opposite direction. The wire PQ is free to slide up and down. To what height will PQ rise? |
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Answer» The magnetic field produced by long straight wire carrying current of 25A rests on a table on small wire `B=(mu_(0)I)/(2pih)` the magnetic force on small conductor is `F=Bilsintheta=Bil` Force applied on PQ balance the weight of small current carrying wire. `F=mg=(mu_(0)I^(2)l)/(2pih)` `h=(mu_(0)I^(2)l)/(2pimg)=(4pixx10^(-7)xx25xx25xx1)/(2pixx2.5xx10^(-3)xx9.8)=51xx10^(-4)` `h=0.51cm` |
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