Chapter 26, Problem 15P

(a)

To determine

The magnetic force on a charge.

Answer to Problem 15P

  $(3.804\times {10}^{-6}\text{N})(-\widehat{k})$

Explanation of Solution

Given:

Charge on the point particle $=q=-3.64\text{ nC =}-3.64\times {10}^{-9}\text{C }$

Velocity of the particle $=\overrightarrow{v}=\left(2750\frac{\text{m}}{\text{s}}\right)\widehat{i}$

Magnetic field $=\overrightarrow{B}=(0.38\text{ T})\widehat{j}$

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 particle is given as

  $\begin{array}{l}\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})\\ \overrightarrow{F}=(-3.64\times {10}^{-9})((2750)\widehat{i}\times (0.38)\widehat{j})\\ \overrightarrow{F}=(-3.64\times {10}^{-9})(2750)(0.38)(\widehat{i}\times \widehat{j})\\ \overrightarrow{F}=(-3.804\times {10}^{-6})(\widehat{k})\\ \overrightarrow{F}=(3.804\times {10}^{-6}\text{N})(-\widehat{k})\end{array}$

Conclusion:

The magnetic force on the particle is $(3.804\times {10}^{-6}N)(-\widehat{k})$ .

(b)

To determine

The magnetic force on a charge.

Answer to Problem 15P

  $(7.51\times {10}^{-6}\text{N})(-\widehat{k})$

Explanation of Solution

Given:

Charge on the point particle $=q=-3.64\text{ nC =}-3.64\times {10}^{-9}\text{C }$

Velocity of the particle $=\overrightarrow{v}=\left(2750\frac{\text{m}}{\text{s}}\right)\widehat{i}$

Magnetic field $=\overrightarrow{B}=(0.75\text{ T})\widehat{i}+(0.75\text{ T})\widehat{j}$

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 particle is given as

  $\begin{array}{l}\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})\\ \overrightarrow{F}=(-3.64\times {10}^{-9})((2750)\widehat{i}\times (0.75\widehat{i}+0.75\widehat{j}))\\ \overrightarrow{F}=(-3.64\times {10}^{-9})(2750)(0.75)(\widehat{i}\times \widehat{i})+(-3.64\times {10}^{-9})(2750)(0.75)(\widehat{i}\times \widehat{j})\\ \overrightarrow{F}=(-3.64\times {10}^{-9})(2750)(0.75)(0)+(-3.64\times {10}^{-9})(2750)(0.75)(\widehat{k})\\ \overrightarrow{F}=0+(-7.51\times {10}^{-6})(\widehat{k})\\ \overrightarrow{F}=(7.51\times {10}^{-6}\text{N})(-\widehat{k})\end{array}$

Conclusion:

The magnetic force on the particle is $(7.51\times {10}^{-6}\text{N})(-\widehat{k})$ .

(c)

To determine

The magnetic force on a charge.

Answer to Problem 15P

  $0\text{ N}$

Explanation of Solution

Given:

Charge on the point particle $=q=-3.64\text{ nC =}-3.64\times {10}^{-9}\text{C }$

Velocity of the particle $=\overrightarrow{v}=\left(2750\frac{\text{m}}{\text{s}}\right)\widehat{i}$

Magnetic field $=\overrightarrow{B}=(0.65\text{ T})\widehat{i}$

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 particle is given as

  $\begin{array}{l}\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})\\ \overrightarrow{F}=(-3.64\times {10}^{-9})((2750)\widehat{i}\times (0.65)\widehat{i})\\ \overrightarrow{F}=(-3.64\times {10}^{-9})(2750)(0.65)(\widehat{i}\times \widehat{i})\\ \overrightarrow{F}=(-3.64\times {10}^{-9})(2750)(0.65)(0)\\ \overrightarrow{F}=0\text{ N}\end{array}$

Conclusion:

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

(d)

To determine

The magnetic force on a charge.

Answer to Problem 15P

  $(7.51\times {10}^{-6}\text{N})(\widehat{j})$

Explanation of Solution

Given:

Charge on the point particle $=q=-3.64\text{ nC =}-3.64\times {10}^{-9}\text{C }$

Velocity of the particle $=\overrightarrow{v}=\left(2750\frac{\text{m}}{\text{s}}\right)\widehat{i}$

Magnetic field $=\overrightarrow{B}=(0.75\text{ T})\widehat{i}+(0.75\text{ T})\widehat{k}$

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 particle is given as

  $\begin{array}{l}\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})\\ \overrightarrow{F}=(-3.64\times {10}^{-9})((2750)\widehat{i}\times (0.75\widehat{i}+0.75\widehat{k}))\\ \overrightarrow{F}=(-3.64\times {10}^{-9})(2750)(0.75)(\widehat{i}\times \widehat{i})+(-3.64\times {10}^{-9})(2750)(0.75)(\widehat{i}\times \widehat{k})\\ \overrightarrow{F}=(-3.64\times {10}^{-9})(2750)(0.75)(0)+(-3.64\times {10}^{-9})(2750)(0.75)(-\widehat{j})\\ \overrightarrow{F}=0+(-7.51\times {10}^{-6})(-\widehat{j})\\ \overrightarrow{F}=(7.51\times {10}^{-6}\text{N})(\widehat{j})\end{array}$

Conclusion:

The magnetic force on the particle is $(7.51\times {10}^{-6}\text{N})(\widehat{j})$ .