A proton,starting from rest,accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040 T at a right angle to the field.What is the radius of the proton's resulting orbit? (e = 1.60 × 10^-19 C,m_proton = 1.67 × 10^-27 kg) A)0.080 m B)0.11 m C)0.14 m D)0.17 m

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Quizplus · Accepted Answer

The radius $r$ of the proton's orbit in a magnetic field can be found using the formula $r = \frac{mv}{qB}$, where $m$ is the mass of the proton, $v$ is the velocity of the proton, $q$ is the charge of the proton, and $B$ is the magnetic field strength. The kinetic energy ($K.E.$) gained by the proton due to acceleration through the potential difference (V) is given by $K.E. = qV$, and since $K.E. = \frac{1}{2}mv^2$, we can solve for $v = \sqrt{\frac{2qV}{m}}$. Substituting $v$ into the radius formula gives $r = \frac{\sqrt{2mV}}{qB}$. Substituting the given values ($m = 1.67 \times 10^{-27}$ kg, $q = 1.60 \times 10^{-19}$ C, $V = 1.0 \times 10^3$ V, and $B = 0.040$ T) into this equation yields a radius close to 0.11 m.

A Proton,starting from Rest,accelerates Through a Potential Difference of 1