Chapter 26, Problem 2STP

To determine

To Choose: The correct option.

Answer to Problem 2STP

The correct option is (A).

Explanation of Solution

Given:

The strength of magnetic field is $B=0.10\text{T}$ .

Radius of the circular path is $r=6.6\text{cm}$.

Formula used:

The electron’s path in the magnetic field is given by

  $Bqv=\frac{m{v}^2}{r}$  ...... (1)

Here, $B$ , $q$ , $m$ , $v$ and $r$ are the strength of magnetic field, charge of the particle, mass of the particle, velocity of the particle, radius of the particles circular path respectively.

Calculation:

Rearranging the equation (1),

  $Bq=\frac{mv}{r}$  ...... (2)

Convert the units for the radius from $\text{cm}$ to $\text{m}$ .

  $\begin{array}{l}r=6.6\text{cm}\\ r=6.6\times \left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\\ r=6.6\times {10}^{-2}\text{m}\end{array}$

Rearrange the equation (2) for $v$ .

  $v=\frac{Bqr}{m}$  ...... (3)

Substitute $1.67\times {10}^{-27}\text{kg}$ for $m$ , $1.6\times {10}^{-19}\text{C}$ for $q$ , $6.6\times {10}^{-2}\text{m}$ for $r$ and 0.10 T for $B$ in equation (3)

  $\begin{array}{l}v=\frac{\left(0.10\text{T}\right)\times \left(1.6\times {10}^{-19}\text{C}\right)\times \left(6.6\times {10}^{-2}\text{m}\right)}{1.67\times {10}^{-27}\text{kg}}\\ \boxed{v=6.3\times {10}^5\text{m/s}}\end{array}$

Conclusion:

Hence, the velocity of the proton in a circular path is $\boxed{6.3\times {10}^5\text{m/s}}$ .