Magnetic Effects Of Current Question 30
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 [CPMT 1988; KCET 1998; UPSEAT 2001]
Options:
A) $ \frac{{{10}^{-19}}}{{\mu_{0}}} $
B) $ {{10}^{-19}}{\mu_{0}} $
C) $ 2\times {{10}^{-10}}{\mu_{0}} $
D) $ \frac{2\times {{10}^{-10}}}{{\mu_{0}}} $
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Answer:
Correct Answer: B
Solution:
$ i=\frac{q}{T}=\frac{2\times 1.6\times {{10}^{-19}}}{2}=1.6\times {{10}^{-19}}A $
$ \therefore ,B=\frac{{\mu_{o}}i}{2r}=\frac{{\mu_{o}}\times 1.6\times {{10}^{-19}}}{2\times 0.8}={\mu_{o}}\times {{10}^{-19}} $
BETA
