Electrons move at right angles to a magnetic | vedclass

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(C) The radius $r$ of the circular path of a charged particle moving perpendicular to a magnetic field is given by the formula $r = \frac{mv}{qB}$.Giv

Vedclass · Accepted Answer

$2.35$

Vedclass · Answer

(C) The radius $r$ of the circular path of a charged particle moving perpendicular to a magnetic field is given by the formula $r = \frac{mv}{qB}$.Given that the specific charge (charge-to-mass ratio) is $\frac{q}{m} = 1.7 	imes 10^{11} 	ext{ C/kg}$,we can rewrite the formula as $r = \frac{v}{(q/m)B}$.Substituting the given values:$v = 6 	imes 10^7 	ext{ m/s}$$\frac{q}{m} = 1.7 	imes 10^{11} 	ext{ C/kg}$$B = 1.5 	imes 10^{-2} 	ext{ T}$$r = \frac{6 	imes 10^7}{(1.7 	imes 10^{11}) 	imes (1.5 	imes 10^{-2})}$$r = \frac{6 	imes 10^7}{2.55 	imes 10^9}$$r \approx 2.35 	imes 10^{-2} 	ext{ m}$Converting to centimeters: $r = 2.35 	ext{ cm}$.

Electrons move at right angles to a magnetic field of $1.5 \times 10^{-2} \text{ T}$ with a speed of $6 \times 10^7 \text{ m/s}$. If the specific charge of the electron is $1.7 \times 10^{11} \text{ C/kg}$,the radius of the circular path will be...... $\text{cm}$.

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