Chapter 26, Problem 53A

To determine

To Write: The speed of a proton moving in present of magnetic field.

Answer to Problem 53A

A proton moves through a $\text{0}\text{.065}\text{T}$ magnetic field strength as shown in the figure below. If the proton moves in a circular path with a radius ofy13y $0.045\text{m}$ , than what is its speed?

Explanation of Solution

Given:

The given expression is

  $q\left(\text{0}\text{.065}\text{T}\right)=\frac{m\left(2.8\times {10}^5\text{m/s}\right)}{0.045\text{m}}$  ...... (1)

Equation (1) is the similar with $rB=\frac{mv}{q}$  ...... (2)

Here, $\text{0}\text{.065}\text{T}$ is magnetic strength ( $B$ ), $2.8\times {10}^5\text{m/s}$ is the velocity ( $v$ ) of the particle and $0.045\text{m}$ is the radius of the circular path.

In this case, the questions should be like: A proton moves through a $\text{0}\text{.065}\text{T}$ magnetic field strength as shown in the figure below. If the proton moves in a circular path with a radius of $0.045\text{m}$ , than what is its speed?

  

Conclusion:

Hence, the required question is given above.