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Question: A helium nucleus makes a full rotation in a circle of radius 0.8 metre in two seconds. The value of ...

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

A

10−19μ0\frac { 10 ^ { - 19 } } { \mu _ { 0 } }

B

10−19μ010 ^ { - 19 } \mu _ { 0 }

C

2×10−10μ02 \times 10 ^ { - 10 } \mu _ { 0 }

D

2×10−10μ0\frac { 2 \times 10 ^ { - 10 } } { \mu _ { 0 } }

Answer

10−19μ010 ^ { - 19 } \mu _ { 0 }

Explanation

Solution

i=qT=2×1.6×10−192=1.6×10−19 Ai = \frac { q } { T } = \frac { 2 \times 1.6 \times 10 ^ { - 19 } } { 2 } = 1.6 \times 10 ^ { - 19 } \mathrm {~A}

∴B=μoi2r=μo×1.6×10−192×0.8=μo×10−19\therefore B = \frac { \mu _ { o } i } { 2 r } = \frac { \mu _ { o } \times 1.6 \times 10 ^ { - 19 } } { 2 \times 0.8 } = \mu _ { o } \times 10 ^ { - 19 }