Potential Difference 52508A

1. **State the problem:** We need to find the potential difference $V$ required to accelerate singly ionized $^{12}C$ atoms so that they move in a circular path of radius $R = 50.0$ cm in a magnetic field $B = 0.150$ T. 2. **Relevant formula:** The radius $R$ of a charged particle moving perpendicular to a magnetic field is given by $$R = \frac{mv}{qB}$$ where $m$ is the mass, $v$ is the velocity, $q$ is the charge, and $B$ is the magnetic field. 3. **Relate velocity to potential difference:** The kinetic energy gained by the ion accelerated through potential difference $V$ is $$\frac{1}{2}mv^2 = qV$$ 4. **Express velocity $v$ from kinetic energy:** $$v = \sqrt{\frac{2qV}{m}}$$ 5. **Substitute $v$ into radius formula:** $$R = \frac{m}{qB} \times \sqrt{\frac{2qV}{m}} = \frac{m}{qB} \times \sqrt{\frac{2qV}{m}}$$ 6. **Simplify the expression:** $$R = \frac{m}{qB} \times \sqrt{\frac{2qV}{m}} = \frac{m}{qB} \times \sqrt{\frac{2qV}{m}} = \frac{m}{qB} \times \sqrt{\frac{2qV}{m}}$$ Rewrite the square root term: $$\sqrt{\frac{2qV}{m}} = \frac{\sqrt{2qV}}{\sqrt{m}}$$ So, $$R = \frac{m}{qB} \times \frac{\sqrt{2qV}}{\sqrt{m}} = \frac{\sqrt{2qVm}}{qB}$$ 7. **Solve for $V$:** $$R = \frac{\sqrt{2qVm}}{qB} \implies RqB = \sqrt{2qVm}$$ Square both sides: $$R^2 q^2 B^2 = 2 q V m$$ Divide both sides by $2 q m$: $$V = \frac{R^2 q B^2}{2 m}$$ 8. **Plug in values:** - $R = 50.0$ cm $= 0.500$ m - $q = 1.602 \times 10^{-19}$ C (charge of singly ionized atom) - $B = 0.150$ T - $m = 1.99 \times 10^{-26}$ kg Calculate: $$V = \frac{(0.500)^2 \times (1.602 \times 10^{-19}) \times (0.150)^2}{2 \times 1.99 \times 10^{-26}}$$ 9. **Calculate numerator:** $$0.500^2 = 0.25$$ $$0.150^2 = 0.0225$$ $$0.25 \times 1.602 \times 10^{-19} \times 0.0225 = 0.25 \times 0.0225 \times 1.602 \times 10^{-19} = 0.005625 \times 1.602 \times 10^{-19} = 9.01 \times 10^{-22}$$ 10. **Calculate denominator:** $$2 \times 1.99 \times 10^{-26} = 3.98 \times 10^{-26}$$ 11. **Calculate $V$:** $$V = \frac{9.01 \times 10^{-22}}{3.98 \times 10^{-26}} = 2.26 \times 10^{4}$$ 12. **Final answer:** $$\boxed{V = 2.26 \times 10^{4} \text{ volts}}$$ This is the potential difference needed for the singly ionized $^{12}C$ atoms to have the specified radius in the magnetic field.