Правка разделов «Solution», «Answer»
en/10.1.1.md
+4 −4
| ### Statement | |||
| $10.1.1.$ A proton accelerated by a voltage of 20 kV enters a uniform magnetic field | |||
| with an induction of 0.1 T perpendicular to the field. Find the radius of the | |||
| circle along which the proton moves in the magnetic field. | |||
| ### Solution | |||
| @@ -9,21 +9,21 @@Solution | |||
| Using: | |||
| \[ | |||
| − | \frac{1}{2}mv^2 = q, \hspace{0.5cm} r=\frac{mv}{qB} | ||
| + | \frac{1}{2}mv^2 = qV, \hspace{0.5cm} r=\frac{mv}{qB} | ||
| \] | |||
| With: | |||
| \[q=1.6 \times 10^{-19}C, \hspace{0.5cm} m=1.67 \times 10^{-27}kg, \hspace{0.5cm} V=20kV, \hspace{0.5cm} B=0.10T\] | |||
| We get: | |||
| \[ | |||
| − | v= \sqrt{\frac{2qV}{m}} \approx 1.96 \times 10^6 \ | ||
| + | v= \sqrt{\frac{2qV}{m}} \approx 1.96 \times 10^6 \hspace{0.12cm} \mathrm{m/s} | ||
| \] | |||
| − | \[r \approx 0.2\mathrm{m} \] | ||
| + | \[r \approx 0.2\hspace{0.12cm}\mathrm{m} \] | ||
| #### Answer | |||
| \[ | |||
| − | \boxed { | ||
| + | \boxed {\hspace{0.12cm} r \approx 0.2m \hspace{0.12cm} } | ||
| \] | |||
| ### Statement | ### Statement | ||
| $10.1.1.$ A proton accelerated by a voltage of 20 kV enters a uniform magnetic field | $10.1.1.$ A proton accelerated by a voltage of 20 kV enters a uniform magnetic field | ||
| with an induction of 0.1 T perpendicular to the field. Find the radius of the | with an induction of 0.1 T perpendicular to the field. Find the radius of the | ||
| circle along which the proton moves in the magnetic field. | circle along which the proton moves in the magnetic field. | ||
| ### Solution | ### Solution | ||
| @@ -9,21 +9,21 @@Solution | |||
| Using: | Using: | ||
| \[ | \[ | ||
| \frac{1}{2}mv^2 = q, \hspace{0.5cm} r=\frac{mv}{qB} | \frac{1}{2}mv^2 = qV, \hspace{0.5cm} r=\frac{mv}{qB} | ||
| \] | \] | ||
| With: | With: | ||
| \[q=1.6 \times 10^{-19}C, \hspace{0.5cm} m=1.67 \times 10^{-27}kg, \hspace{0.5cm} V=20kV, \hspace{0.5cm} B=0.10T\] | \[q=1.6 \times 10^{-19}C, \hspace{0.5cm} m=1.67 \times 10^{-27}kg, \hspace{0.5cm} V=20kV, \hspace{0.5cm} B=0.10T\] | ||
| We get: | We get: | ||
| \[ | \[ | ||
| v= \sqrt{\frac{2qV}{m}} \approx 1.96 \times 10^6 \ |
v= \sqrt{\frac{2qV}{m}} \approx 1.96 \times 10^6 \hspace{0.12cm} \mathrm{m/s} | ||
| \] | \] | ||
| \[r \approx 0.2\mathrm{m} \] | \[r \approx 0.2\hspace{0.12cm}\mathrm{m} \] | ||
| #### Answer | #### Answer | ||
| \[ | \[ | ||
| \boxed { |
\boxed {\hspace{0.12cm} r \approx 0.2m \hspace{0.12cm} } | ||
| \] | \] | ||