A beam of neutrons performs circular motion o | vedclass

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(C) Given: radius $r = 1 \, m$,inhomogeneity $\Delta r = 0.01 \, m$,speed $v = 54 \, m/s$,magnetic moment $M = 9.67 	imes 10^{-27} \, J/T$,and mass $

Vedclass · Accepted Answer

$5.04$

Vedclass · Answer

(C) Given: radius $r = 1 \, m$,inhomogeneity $\Delta r = 0.01 \, m$,speed $v = 54 \, m/s$,magnetic moment $M = 9.67 	imes 10^{-27} \, J/T$,and mass $m = 1.67 	imes 10^{-27} \, kg$.The magnetic force experienced by the neutron in an inhomogeneous magnetic field is given by $F = M \frac{\Delta B}{\Delta r}$.Since the neutron is performing circular motion,this magnetic force provides the necessary centripetal force: $F = \frac{m v^2}{r}$.Equating the two expressions: $M \frac{\Delta B}{\Delta r} = \frac{m v^2}{r}$.Rearranging to solve for the variation in magnetic field $\Delta B$: $\Delta B = \frac{m v^2 \Delta r}{M r}$.Substituting the given values: $\Delta B = \frac{1.67 	imes 10^{-27} 	imes (54)^2 	imes 0.01}{9.67 	imes 10^{-27} 	imes 1}$.$\Delta B = \frac{1.67 	imes 2916 	imes 0.01}{9.67} \approx \frac{48.6972}{9.67} \approx 5.04 \, T$.

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