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If a long straight wire carries a current of 40 A, then the magnitude of the field B at a point 15 cm away from the wire isA. `5.34xx10^(-5)T`B. `8.34xx10^(-5)T`C. `9.6xx10^(-5)T`D. `10.2xx10^(-5)T` |
Answer» Correct Answer - A I = 40 A r = 15 cm `=15xx10^(-2)m` `therefore" "B=(mu_(0)I_("enc"))/(2piR)=2xx10^(-7)xx(75)/(3)=5xx10^(-6)T` |
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