Chapter 20, Problem 23P

To determine

The deflection distance of bullet from its path.

Answer to Problem 23P

  $6.05\times {10}^{-7}\text{ m}$

Explanation of Solution

Given:

Mass $m=3.4$ g $=3.4\times {10}^{-3}$ kg

Speed $v=160$ m/s

Earth’s magnetic field $B=5\times {10}^{-5}$ T

Charge $q=13.5\times {10}^{-9}$ C

Distance $d=1$ km $=1\times {10}^3$ km

Formula used:

Magnetic force

  $F_m=qvB$

Where,

  $q=$ Charge

  $v=$ Velocity

  $B=$ Magnetic Field

Calculation:

Magnetic force on the bullet,

  $\begin{array}{l}{F}_m=qvB\\ F_m=13.5\times {10}^{-9}\times 160\times 5\times {10}^{-5}\\ F_m=1.08\times {10}^{-10}\text{ N}\end{array}$

Now, acceleration of bullet,

  $\begin{array}{l}{F}_m=ma\\ a=\frac{F_m}{m}\\ a=\frac{1.08\times {10}^{-10}}{3.4\times {10}^{-3}}\\ a=0.31\times {10}^{-7}{\text{ m/s}}^2\end{array}$

Time taken by bullet,

  $\begin{array}{l}t=\frac{d}{v}\\ t=\frac{1\times {10}^3}{160}\\ t=6.25\sec \end{array}$

Consider that the bullet is initially at rest, so the initial speed is zero.

By the Newton’s law of motion,

  $\begin{array}{l}s=ut+\frac{1}{2}a{t}^2\\ s=0\times 6.25+\frac{1}{2}\times 0.31\times {10}^{-7}\times {\left(6.25\right)}^2\\ s=6.05\times {10}^{-7}\text{ m}\end{array}$

Conclusion:

Therefore, the deflection of bullet is equal to the $6.05\times {10}^{-7}\text{ m}$ .