Moving Charges and Magnetism

Question: A helium nucleus makes a full rotation in a circle of radius 0.8 metre in two seconds. The value of the magnetic field B at the centre of the circle will be

1) $\frac{{{{10}^{ - 19}}}}{{{\mu _0}}}$

2) ${10^{ - 19}}{\mu _0}$

3) $2 \times {10^{ - 10}}{\mu _0}$

4) $\frac{{2 \times {{10}^{ - 10}}}}{{{\mu _0}}}$

Solution:

Explanation: No Explanation

Biot-Savart's Law Amperes Law

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