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A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field B. What is the magnitude of the magnetic field? A. `2`B. `1*5`C. `0*55`D. `0*65` |
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Answer» Correct Answer - D Here, `m=200g=200xx10^-3kg`, `l=1*5m`, `I=2A`, `g=9*8m//s^2`, `theta=90^@` Force on current carrying conductor in mag. field is given by `F=IlB sin theta=IlBsin 90^@=IlB` Weight `W=mg=200xx10^-3xx9*8N` As it is suspended in mid air, So `W=F` `mg=IlB` or `B=(mg)/(Il)=(200xx10^-3xx9*8)/(2xx1*5)=0*65T` |
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