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$(C)$ The charge on a helium nucleus is $q = 2e = 2 	imes 1.6 	imes 10^{-19} \,C = 3.2 	imes 10^{-19} \,C$. Given radius $r = 0.8 \,m$ and time per

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$4 \pi 	imes 10^{-26} \,T$

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$(C)$ The charge on a helium nucleus is $q = 2e = 2 	imes 1.6 	imes 10^{-19} \,C = 3.2 	imes 10^{-19} \,C$. Given radius $r = 0.8 \,m$ and time period $T = 2.5 \,s$. The equivalent current $I$ due to the rotation of the charge is $I = \frac{q}{T} = \frac{3.2 	imes 10^{-19}}{2.5} = 1.28 	imes 10^{-19} \,A$. The magnetic field $B$ at the centre of a circular loop is given by $B = \frac{\mu_0 I}{2r}$. Substituting the values: $B = \frac{4 \pi 	imes 10^{-7} 	imes 1.28 	imes 10^{-19}}{2 	imes 0.8} = \frac{4 \pi 	imes 10^{-7} 	imes 1.28}{1.6} = 4 \pi 	imes 10^{-7} 	imes 0.8 	imes 10^{-19} = 3.2 \pi 	imes 10^{-26} \,T$. Rounding to the nearest provided option, the value is $4 \pi 	imes 10^{-26} \,T$.

When a helium nucleus makes a full rotation of a circle of radius $0.8 \,m$ in $2.5 \,s$, then the value of magnetic field $B$ at the centre of the circle will be

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