Chapter 29, Problem 78A

To determine

To calculate the radius of curvature of a helium particle in a uniform magnetic field of $6.5\times {10}^{-2}$ T.

Answer to Problem 78A

The radius of curvature is 0.010 m

Explanation of Solution

Given data:

Mass of doubly ionized helium atom ( M ) = $6.7\times {10}^{-27}$ kg

Potential difference ( V ) = 1.0 kV

Magnetic field ( B ) = $6.5\times {10}^{-2}$ T

Charge of electron ( q ) = $1.6\times {10}^{-19}$ C

For 2 electrons = $2\left(1.6\times {10}^{-19}\right)$ C

Formula used:

The radius ( R ) of the curvature,

  $R=\frac{1}{B}\sqrt{\frac{2\left(V\right)\left(M\right)}{q}}$

Calculation:

The radius ( R ) of the curvature,

  $\begin{array}{l}R=\frac{1}{B}\sqrt{\frac{2\left(V\right)\left(M\right)}{q}}=\frac{1}{6.5\times {10}^{-2}}\sqrt{\frac{2\left(1\right)\left(6.7\times {10}^{-27}\right)}{2\left(1.6\times {10}^{-19}\right)}}\\ R=0.010\text{ m}\end{array}$

Conclusion:

Hence, the radius of curvature is 0.010 m.