Chapter 26, Problem 40A

To determine

To rank: the particles which enter regions with different magnetic field along with different speed.

Answer to Problem 40A

  $\boxed{r_3=16.67\times {10}^3\text{m}<{\left(r_4=r\right)}_1=33.33\times {10}^3\text{m}\text{<}{r}_5=333.3\times {10}^3\text{m}<r_2=666.67\times {10}^3\text{m}}$ .

Explanation of Solution

Given:

There are 5 regions where magnetic field strength and speed of the particles are different.

First region:

Magnetic field is $B=0.12\text{T}$ and speed of the particle is $v=4\times {10}^3\text{m/s}$.

Second region:

Magnetic field is $B=0.12\text{T}$ and speed of the particle is $v=8\times {10}^4\text{m/s}$.

Third region:

Magnetic field is $B=0.24\text{T}$ and speed of the particle is $v=4\times {10}^3\text{m/s}$.

Fourth region:

Magnetic field is $B=0.12\text{T}$ and speed of the particle is $v=4\times {10}^3\text{m/s}$.

Fifth region:

Magnetic field is $B=0.24\text{T}$ and speed of the particle is $v=8\times {10}^4\text{m/s}$.

Formula used:

The expression for radius of the circular path traversed by the particle is

  $r=\frac{mv}{qB}$  ...... (1)

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

Calculation:

Since the particle is same in the entire five regions, so in equation (1) mass $\left(m\right)$ and charge $\left(q\right)$ will remain the same, so these two terms are ignored from the equation for simplification.

  $r=\frac{v}{B}$  ...... (2)

For first region, substitute $4\times {10}^3\text{m/s}$ for $v$ and $0.12\text{T}$ for $B$ in equation (1)

  $\begin{array}{l}{r}_1=\frac{4\times {10}^3\text{m/s}}{0.12\text{T}}\\ \boxed{r_1=33.33\times {10}^3\text{m}}\end{array}$

For second region, substitute $8\times {10}^4\text{m/s}$ for $v$ and $0.12\text{T}$ for $B$ in equation (1)

  $\begin{array}{l}{r}_2=\frac{8\times {10}^4\text{m/s}}{0.12\text{T}}\\ r_2=66.66\times {10}^4\text{m}\\ \boxed{r_2=666.67\times {10}^3\text{m}}\end{array}$

For third region, substitute $4\times {10}^3\text{m/s}$ for $v$ and $0.24\text{T}$ for $B$ in equation (1)

  $\begin{array}{l}{r}_3=\frac{4\times {10}^3\text{m/s}}{0.24\text{T}}\\ \boxed{r_3=16.67\times {10}^3\text{m}}\end{array}$

For fourth region, substitute $4\times {10}^3\text{m/s}$ for $v$ and $0.12\text{T}$ for $B$ in equation (1)

  $\begin{array}{l}{r}_4=\frac{4\times {10}^3\text{m/s}}{0.12\text{T}}\\ \boxed{r_4{}_{}=33.33\times {10}^3\text{m}}\end{array}$

For fifth region, substitute $8\times {10}^4\text{m/s}$ for $v$ and $0.24\text{T}$ for $B$ in equation (1)

  $\begin{array}{l}{r}_5=\frac{8\times {10}^4\text{m/s}}{0.24\text{T}}\\ \boxed{r_5=333.3\times {10}^3\text{m}}\end{array}$

Therefore, from the above calculation, it is found that, radius will be larger for second region but first and fourth region, radius is equal in magnitude. They are arranged below in ascending order

  $\boxed{r_3=16.67\times {10}^3\text{m}<{\left(r_4=r\right)}_1=33.33\times {10}^3\text{m}\text{<}{r}_5=333.3\times {10}^3\text{m}<r_2=666.67\times {10}^3\text{m}}$ .

Conclusion:

Hence, the order is third region < first region = fourth region< fifth region< second region