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Due to `10` ampere of current flowing in a circular coil of `10 cm` radius, the magnetic field produced at its centre is `3.14xx10^(-3) Weber//m^(3)`. The number of turns in the coil will beA. `5000`B. `100`C. `50`D. `25` |
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Answer» Correct Answer - C `B=(mu_(0))/(4pi).(2piNi)/r` `implies 3.14x10^(-3)=(10^(-7)xx2xx3.14xxNxx10)/((10xx10^(-2)) implies N=50` |
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