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A helium nucleus makes full rotation in a circle of radius 0.8 m in 2.5 seconds. The value of magnetic field B at the centre of the circle will beA. `3.1pi xx10^-26 T`B. `2pi xx10^-26 T`C. `4pi xx10^-25 T`D. `2 pi xx10^-25 T` |
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Answer» Correct Answer - A `B=(mu_0I)/(2a)(I=(q)/(T)=(2e)/(T))` Since for helium , `q=2e` `B=(mu_02e)/(2aT)=(4pi xx 10^-6xx2xx1.6xx10^-19)/(2xx0.8xx2.5)` `B=3.2pi xx10^-26 T`. |
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