Chapter 29, Problem 29.10P

A laboratory electromagnet produces a magnetic field of magnitude 1.50 T. A proton moves through this field with a speed of 6.00 × 10⁶ m/s. (a) Find the magnitude of the maximum magnetic force that could he exerted on the proton. (b) What is the magnitude of the maximum acceleration of the proton? (c) Would the field exert the same magnetic force on an electron moving through the field with the same speed? (d) Would the electron experience the same acceleration? Explain.

To determine

The maximum magnetic force that is exerted on the proton.

Answer to Problem 29.10P

The maximum magnetic force that is exerted on the proton is $14.4\times {10}^{-13}\text{N}$.

Explanation of Solution

Given info: A magnetic field of magnitude $1.50\text{T}$ is produced by the laboratory electromagnet and a proton moves through the field with the speed of $6\times {10}^6\text{m}/\text{s}$.

Explanation:

The formula to calculate the Magnetic force acting on a moving charge particle is,

$F=q\left(v\times B\right)$

Here,

$q$ is the charge of the particle.

$v$ is the velocity of the particle.

$B$ is the magnetic field.

The cross product of $v\times B$ is expanded by the formula as,

$v\times B=vB\mathrm{Sin}\theta$

Here,

$\mathrm{Sin}\theta$ is the vertical component between $v$ and $B$.

Substitute $vB\mathrm{Sin}\theta$ for $v\times B$ in the magnetic force formula,

$F=qvB\mathrm{Sin}\theta$

The cross product is maximum for $\theta =90°$ then the above formula becomes,

$F=qvB$

The mass of proton is $1.67\times {10}^{-27}\text{Kg}$ and charge of proton is $1.6\times {10}^{-19}\text{C}$

Substitute $1.6\times {10}^{-19}\text{C}$ for $q$, $6\times {10}^6\text{m}/\text{s}$ for $v$, $1.50\text{T}$ for $B$ in the above formula,

$\begin{array}{c}F=qvB\\ =\left(1.6\times {10}^{-19}\text{C}\right)\left(6\times {10}^6\text{m}/\text{s}\right)\left(1.50\text{T}\right)\\ =14.4\times {10}^{-13}\text{N}\end{array}$

Conclusion:

Therefore, the maximum magnetic force that is exerted on the proton is $14.4\times {10}^{-13}\text{N}$

To determine

 The magnitude of maximum acceleration of the proton.

Answer to Problem 29.10P

The magnitude of maximum acceleration of the proton is $86.22\times {10}^{13}\text{m}/{\text{s}}^{\text{2}}$

Explanation of Solution

Given info: A magnetic field of magnitude $1.50\text{T}$ is produced by the laboratory electromagnet and a proton moves through the field with the speed of $6\times {10}^6\text{m}/\text{s}$.

Explanation:

The formula to calculate the force acting on a proton of mass $m$ is,

$F=ma$

Here,

$m$ is the mass of proton.

$a$ is the proton’s acceleration.

Substitute $1.67\times {10}^{-27}\text{Kg}$ for $q$, $14.4\times {10}^{-13}\text{N}$ for $F$, in the above formula,

$\begin{array}{c}F=ma\\ \left(14.4\times {10}^{-13}\text{N}\right)=\left(1.67\times {10}^{-27}\text{Kg}\right)a\\ a=\frac{14.4\times {10}^{-13}}{1.67\times {10}^{-27}}\\ =86.22\times {10}^{13}\text{m}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the magnitude of maximum acceleration of the proton is $86.22\times {10}^{13}\text{m}/{\text{s}}^{\text{2}}$

To determine

Whether the maximum magnetic force that is exerted on the proton is same or not for the electron.

Answer to Problem 29.10P

The maximum magnetic force that is exerted in the case of proton is same for electron.

Explanation of Solution

Given info: A magnetic field of magnitude $1.50\text{T}$ is produced by the laboratory electromagnet and a electron moves through the field with the speed of $6\times {10}^6\text{m}/\text{s}$.

Explanation:

The formula to calculate the Magnetic force acting on a moving charge particle is,

$F=q\left(v\times B\right)$

Here,

$q$ is the charge of the particle.

$v$ is the velocity of the particle.

$B$ is the magnetic field.

The cross product of $v\times B$ is expanded by the formula as,

$v\times B=vB\mathrm{Sin}\theta$

Here,

$\mathrm{Sin}\theta$ is the vertical component between $v$ and $B$.

Substitute $vB\mathrm{Sin}\theta$ for $v\times B$ in the magnetic force formula,

$F=qvB\mathrm{Sin}\theta$

The cross product is maximum for $\theta =90°$ above formula becomes,

$F=qvB$

The mass of electron is $9.11\times {10}^{-31}\text{Kg}$ and charge of electron is $1.6\times {10}^{-19}\text{C}$.

Substitute $1.6\times {10}^{-19}\text{C}$ for $q$, $6\times {10}^6\text{m}/\text{s}$ for $v$, $1.50T$ for $B$ in the above formula,

$\begin{array}{c}F=qvB\\ =\left(1.6\times {10}^{-19}\text{C}\right)\left(6\times {10}^6\text{m}/\text{s}\right)\left(1.50\text{T}\right)\\ =14.4\times {10}^{-13}\text{N}\end{array}$

Conclusion:

Therefore, the maximum magnetic force that is exerted in the case of proton is same for the electron.

To determine

Whether the magnitude of maximum acceleration of the proton is same for electron or not.

Answer to Problem 29.10P

The magnitude of maximum acceleration obtained in the case of proton is not same for the electron.

Explanation of Solution

Given info: The proton makes an angle of $60°$ with the direction of a magnetic field of magnitude $0.180\text{T}$ in the positive $x-$ direction moving with the velocity of $5.02\times {10}^6\text{m}/\text{s}$.

Explanation:

The formula to calculate the force acting on a proton of mass $m$ is,

$F=ma$

Here,

$m$ is the mass of proton.

$a$ is the proton’s acceleration.

Substitute $9.11\times {10}^{-31}\text{Kg}$ for $m$, $1.25\times {10}^{-13}\text{N}$ for $F$, in the above formula,

$\begin{array}{c}F=ma\\ \left(14.4\times {10}^{-13}\text{N}\right)=\left(9.11\times {10}^{-31}\text{Kg}\right)a\\ a=\frac{14.4\times {10}^{-13}}{9.1\times {10}^{-31}}\\ =15.82\times {10}^{17}\text{m}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the magnitude of maximum acceleration obtained in the case of proton is not same for the electron.