A charge having q/m equal to 10^8 , C/kg and | vedclass

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(B) The radius of the helical path of a charged particle moving in a uniform magnetic field is given by the formula:$r = \frac{mv_{\perp}}{qB} = \frac

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$0.5$

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(B) The radius of the helical path of a charged particle moving in a uniform magnetic field is given by the formula:$r = \frac{mv_{\perp}}{qB} = \frac{mv \sin 	heta}{qB}$Given values:$q/m = 10^8 \, C/kg \implies m/q = 10^{-8} \, kg/C$$v = 3 	imes 10^5 \, m/s$$B = 0.3 \, T$$	heta = 30^{\circ}$Substituting the values into the formula:$r = \left( \frac{m}{q} ight) \frac{v \sin 30^{\circ}}{B}$$r = (10^{-8}) 	imes \frac{3 	imes 10^5 	imes 0.5}{0.3}$$r = 10^{-8} 	imes \frac{1.5 	imes 10^5}{0.3}$$r = 10^{-8} 	imes 5 	imes 10^5 = 5 	imes 10^{-3} \, m$Converting meters to centimeters:$r = 5 	imes 10^{-3} 	imes 10^2 \, cm = 0.5 \, cm$.

$A$ charge having $q/m$ equal to $10^8 \, C/kg$ and with velocity $3 \times 10^5 \, m/s$ enters into a uniform magnetic field $0.3 \, T$ at an angle $30^{\circ}$ with the direction of the field. The radius of curvature will be ...... $cm$.

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