Question

A helium nucleus makes a full rotation in a circle of radius 0.8 m in 2s. The value of the magnetic field at the centre of the circle will be

• A. $ {{\mu }_{0}}\,\times \,{{10}^{-19}}\,T $
• B. $ 1.6\,\times {{\mu }_{0}}\,\times \,{{10}^{-19}}\,T $
• C. $ 3{{\mu }_{0}}\,\times \,{{10}^{-19}}\,T $
• D. $ 2{{\mu }_{0}}\,\times \,{{10}^{-19}}\,T $

Answer 1

Answer: (A)

The magnetic field at centre
$ B=\frac{{{\mu }_{0}}I}{2r}=\frac{{{\mu }_{0}}(q-t)}{2r} $
$ =\frac{{{\mu }_{0}}\times 2\times 1.6\times {{10}^{-19}}}{2\times 0.8\times 2} $
$ B={{\mu }_{0}}\times {{10}^{-19}}T $