Question
A helium nucleus makes a full rotation in a circle of radius $0.8 \mathrm{~m}$ in two seconds. The value of the magnetic field $B$ at centre of circle will be :(a) $\frac{10^{-19}}{\mu_{0}}$(b) $10^{-19} \mu_{0}$(c) $2 \times 10^{-19} \mu_{0}$(d) $\frac{2 \times 10^{19}}{\mu_{0}}$
Step 1
8 \, \mathrm{m}$ in $T = 2 \, \mathrm{s}$. We need to find the magnetic field at the center of the circle. Show more…
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A He nucleus makes a full rotation in a circle of radius $0.8$ meter in $2 \mathrm{sec}$. The value of the mag. field $\mathrm{B}$ at the centre of the circle will be $\quad$ Tesla. (a) $\left(10^{-19} / \mu_{0}\right)$ (b) $10^{-19} \mu_{0}$ (c) $2 \times 10^{-10} \mathrm{H}_{0}$ (d) $\left[\left(2 \times 10^{-10}\right) / \mu_{0}\right]$
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Round 2
A helium nucleus consists of two protons (charge $e=$ $1.60 \times 10^{-19} \mathrm{C}$ ) and two neutrons (zero charge). Assuming the protons are separated by $1.9 \times 10^{-15} \mathrm{~m},$ the electric potential en- ergy is (a) $4.8 \times 10^{-13} \mathrm{~J} ;$ (b) $2.4 \times 10^{-19} \mathrm{~J} ;$ (c) $3.4 \times 10^{-11} \mathrm{~J}$ (d) $1.2 \times 10^{-13} \mathrm{~J}$
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