An ion beam of specific charge 5 times 10^7 | vedclass

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(A) Given: Specific charge $\frac{q}{m} = 5 	imes 10^7 \ C/kg$,Magnetic field $B = 4 	imes 10^{-2} \ T$,Velocity $v = 2 	imes 10^5 \ m/s$.When a ch

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(A) Given: Specific charge $\frac{q}{m} = 5 	imes 10^7 \ C/kg$,Magnetic field $B = 4 	imes 10^{-2} \ T$,Velocity $v = 2 	imes 10^5 \ m/s$.When a charged particle enters a uniform magnetic field perpendicularly,it follows a circular path.The radius $r$ of the circular path is given by the formula:$r = \frac{mv}{qB} = \frac{v}{(\frac{q}{m})B}$Substituting the given values:$r = \frac{2 	imes 10^5}{(5 	imes 10^7) 	imes (4 	imes 10^{-2})}$$r = \frac{2 	imes 10^5}{20 	imes 10^5}$$r = \frac{2}{20} = 0.1 \ m$Therefore,the radius of the circular path is $0.1 \ m$.

An ion beam of specific charge $5 \times 10^7 \ C/kg$ enters a uniform magnetic field of $4 \times 10^{-2} \ T$ with a velocity $2 \times 10^5 \ m/s$ perpendicularly. The radius of the circular path of the ions in meters will be:

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