Chapter 26, Problem 16P

(a)

To determine

The magnetic force on a proton.

Answer to Problem 16P

  $(6.4\times {10}^{-16}N)(-\widehat{j})$

Explanation of Solution

Given:

Magnitude of charge on proton $=q=1.6\times {10}^{-19}\text{C}$

Velocity of the proton $=\overrightarrow{v}=\left(2.7\frac{\text{km}}{\text{s}}\right)\widehat{i}=\left(2.7\times {10}^3\frac{\text{m}}{\text{s}}\right)\widehat{i}$

Magnetic field $=\overrightarrow{B}=(1.48\text{ T})\widehat{k}$

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

  $\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$

Calculation:

Magnetic force on the proton is given as

  $\begin{array}{l}\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})\\ \overrightarrow{F}=(1.6\times {10}^{-19})((2.7\times {10}^3)\widehat{i}\times (1.48)\widehat{k})\\ \overrightarrow{F}=(1.6\times {10}^{-19})(2.7\times {10}^3)(1.48)(\widehat{i}\times \widehat{k})\\ \overrightarrow{F}=(6.4\times {10}^{-16})(-\widehat{j})\\ \overrightarrow{F}=(6.4\times {10}^{-16}\text{N})(-\widehat{j})\end{array}$

Conclusion:

The magnetic force on the proton is $(6.4\times {10}^{-16}\text{N})(-\widehat{j})$ .

(b)

To determine

The magnetic force on a proton.

Answer to Problem 16P

  $(8.76\times {10}^{-16}\text{N})(-\widehat{j})$

Explanation of Solution

Given:

Magnitude of charge on proton $=q=1.6\times {10}^{-19}\text{C}$

Velocity of the proton $=\overrightarrow{v}=\left(3.7\frac{\text{km}}{\text{s}}\right)\widehat{j}=\left(3.7\times {10}^3\frac{\text{m}}{\text{s}}\right)\widehat{j}$

Magnetic field $=\overrightarrow{B}=(1.48\text{ T})\widehat{k}$

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

  $\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$

Calculation:

Magnetic force on the proton is given as

  $\begin{array}{l}\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})\\ \overrightarrow{F}=(1.6\times {10}^{-19})((3.7\times {10}^3)\widehat{j}\times (1.48)\widehat{k})\\ \overrightarrow{F}=(1.6\times {10}^{-19})(3.7\times {10}^3)(1.48)(\widehat{j}\times \widehat{k})\\ \overrightarrow{F}=(8.76\times {10}^{-16})(\widehat{i})\\ \overrightarrow{F}=(8.76\times {10}^{-16}\text{N})(-\widehat{j})\end{array}$

Conclusion:

The magnetic force on the proton is $(8.76\times {10}^{-16}\text{N})(-\widehat{j})$ .

(c)

To determine

The magnetic force on a proton.

Answer to Problem 16P

  $0\text{ N}$

Explanation of Solution

Given:

Magnitude of charge on proton $=q=1.6\times {10}^{-19}\text{C}$

Velocity of the proton $=\overrightarrow{v}=\left(6.8\frac{\text{km}}{\text{s}}\right)\widehat{k}=\left(6.8\times {10}^3\frac{\text{m}}{\text{s}}\right)\widehat{k}$

Magnetic field $=\overrightarrow{B}=(1.48\text{ T})\widehat{k}$

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

  $\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$

Calculation:

Magnetic force on the proton is given as

  $\begin{array}{l}\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})\\ \overrightarrow{F}=(1.6\times {10}^{-19})((6.8\times {10}^3)\widehat{k}\times (1.48)\widehat{k})\\ \overrightarrow{F}=(1.6\times {10}^{-19})(6.8\times {10}^3)(1.48)(\widehat{k}\times \widehat{k})\\ \overrightarrow{F}=(8.76\times {10}^{-16})(0)\\ \overrightarrow{F}=0\text{ N}\end{array}$

Conclusion:

The magnetic force on the proton is $0\text{ N}$ .

(d)

To determine

The magnetic force on a proton.

Answer to Problem 16P

  $(7.10\times {10}^{-16})\widehat{i}-(9.47\times {10}^{-16})\widehat{j}$

Explanation of Solution

Given:

Magnitude of charge on proton $=q=1.6\times {10}^{-19}\text{C}$

Velocity of the proton $=\overrightarrow{v}=\left(4\frac{\text{km}}{\text{s}}\right)\widehat{i}+\left(3\frac{\text{km}}{\text{s}}\right)\widehat{j}=\left(4\times {10}^3\frac{\text{m}}{\text{s}}\right)\widehat{i}+\left(3\times {10}^3\frac{\text{m}}{\text{s}}\right)\widehat{j}$

Magnetic field $=\overrightarrow{B}=(1.48\text{ T})\widehat{k}$

Formula Used:

Magnetic force on a moving charged particle in a magnetic field region is given as

  $\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$

Calculation:

Magnetic force on the proton is given as

  $\begin{array}{l}\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})\\ \overrightarrow{F}=(1.6\times {10}^{-19})(((4\times {10}^3)\widehat{i}+(3\times {10}^3)\widehat{j})\times (1.48)\widehat{k})\\ \overrightarrow{F}=(1.6\times {10}^{-19})(4\times {10}^3)(1.48)(\widehat{i}\times \widehat{k})+(1.6\times {10}^{-19})(3\times {10}^3)(1.48)(\widehat{j}\times \widehat{k})\\ \overrightarrow{F}=(9.47\times {10}^{-16})(-\widehat{j})+(7.10\times {10}^{-16})(\widehat{i})\\ \overrightarrow{F}=(7.10\times {10}^{-16})\widehat{i}-(9.47\times {10}^{-16})\widehat{j}\end{array}$

Conclusion:

The magnetic force on the proton is $(7.10\times {10}^{-16})\widehat{i}-(9.47\times {10}^{-16})\widehat{j}$ .