A helium nucleus make 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?
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Answered by
3
Answer:
The current due to the revolution of the He nucleus is I=
2π
qω
=
2
2×1.6×10
−19
=1.6×10
−19
A
Magnetic field at the center B=
2r
μ
0
I
=
2×0.8
μ
0
×1.6×10
−19
=10
−19
μ
0
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Answered by
1
Answer:
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