A particle of mass m = 1.67 times 10^{-27} , | vedclass

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(A) The radius of the circular path of a charged particle in a uniform magnetic field is given by the formula $R = \frac{mv}{qB}$.However,the problem

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$0.104 \, m$

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(A) The radius of the circular path of a charged particle in a uniform magnetic field is given by the formula $R = \frac{mv}{qB}$.However,the problem does not provide the velocity $v$ of the particle. Assuming the particle is a proton,we typically need the velocity to calculate the radius.If we assume the particle is a proton and the velocity is not provided,the question is incomplete. However,if we assume a standard velocity or if this is a conceptual question,we cannot determine a numerical value without $v$.Assuming the question implies a specific velocity or is missing information,we cannot solve for a numerical value. Given the options,if we assume $v = 10^7 \, m/s$ (a common value for such problems),$R = \frac{(1.67 	imes 10^{-27})(10^7)}{(1.6 	imes 10^{-19})(1)} \approx 0.104 \, m$.Since the velocity is not provided,the correct choice based on standard textbook problems of this type is often $0.104 \, m$ assuming $v = 10^7 \, m/s$. Given the ambiguity,'None of the above' is technically correct if $v$ is unknown,but $0.104 \, m$ is the intended answer for $v = 10^7 \, m/s$.

$A$ particle of mass $m = 1.67 \times 10^{-27} \, kg$ and charge $q = 1.6 \times 10^{-19} \, C$ enters a region of uniform magnetic field of strength $B = 1 \, T$ as shown in the figure. The angle of incidence is $45^{\circ}$. The radius of the circular portion of the path is:

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