Chapter 14, Problem 2SP

A small metal ball with a charge of +0.08 C and a mass of 25 g (0.025 kg) enters a region where there is a magnetic field of 0.5 T. The ball is traveling with a velocity of 110 m/s in a direction perpendicular to the magnetic field, as shown in this problem's diagram.

a.    What is the magnitude of the magnetic force acting on the ball?

b.    What is the direction of the magnetic force exerted on the ball when it is at the position shown?

c.    Will this force change the magnitude of the velocity of the ball? Explain.

d.    From Newton’s second law, what is the magnitude of the acceleration of the charged hall?

e.    Because centripetal acceleration is equal to v²/r, what is the radius of the curve the ball will move through under the influence of the magnetic force?

To determine

The magnitude of magnetic force on the ball.

Answer to Problem 2SP

Magnetic force on the ball is $4.4\text{N}$.

Explanation of Solution

Given Info The charge of metal ball is $+0.08\text{C}$, mass is $0.025\text{kg}$, magnetic field is $0.5\text{T}$ and velocity of ball is $110\text{m}/\text{s}$.

Write the equation for Lorentz magnetic force acting on ball moves in direction perpendicular to magnetic field.

$F=qvB$

Here,

$F$ is the magnetic force

$q$ is the charge

$v$ is the velocity of ball

$B$ is the magnetic field

Substitute $+0.08\text{C}$ for $q$,$110\text{m}/\text{s}$ for $v$ and $0.5\text{T}$ for $B$ in the above equation to find $F$.

$\begin{array}{c}F=\left(+0.08\text{C}\right)\left(110\text{m}/\text{s}\right)\left(0.5\text{T}\right)\\ =4.4\text{N}\end{array}$

Conclusion:

Therefore, the force is $4.4\text{N}$.

To determine

The direction of magnetic force on the ball.

Answer to Problem 2SP

Magnetic force acts into the page.

Explanation of Solution

Direction of magnetic force on ball is given by the right hand thumb rule. If the index finger denotes the direction of motion of ball, middle finger represents the direction of magnetic field; right hand thumb represents the direction of magnetic force. Force which is perpendicular to both velocity and magnetic field points into the page.

Conclusion:

Therefore, the Magnetic force acts into the page.

To determine

To Check: Whether the force on ball can change the velocity of ball or not.

Answer to Problem 2SP

Yes. Force on ball can change the velocity of ball.

Explanation of Solution

Magnetic force acting on the ball is perpendicular to both the direction of motion and magnetic field. This force put the ball in circular motion. The necessary centripetal force is provided by the magnetic force calculated in part (a).

In a circular motion, direction of motion is changing at every instant. This means even the speed of ball is constant, there will be a change in velocity.

Conclusion:

Therefore, the force on ball can change the velocity of ball.

To determine

The acceleration of charged ball.

Answer to Problem 2SP

The acceleration of ball is $176\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Given Info The charge of metal ball is $+0.08\text{C}$, mass is $0.025\text{kg}$, magnetic field is $0.5\text{T}$ and velocity of ball is $110\text{m}/\text{s}$.

Write the formula to calculate the acceleration of ball.

$a=\frac{F}{m}$

Here,

$a$ is the centripetal acceleration of ball

$m$ is the mass  of ball

Substitute $4.4\text{N}$ for $F$ and $0.025\text{kg}$ for $m$ in the above equation to find $a$.

$\begin{array}{c}a=\frac{4.4\text{N}}{0.025\text{kg}}\\ =176\text{m}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the acceleration of ball is $176\text{m}/{\text{s}}^{\text{2}}$.

To determine

The radius of circular path which ball moves under the influence of magnetic field.

Answer to Problem 2SP

The radius of circular path is $68.75\text{m}$,

Explanation of Solution

Given Info The charge of metal ball is $+0.08\text{C}$, mass is $0.025\text{kg}$, magnetic field is $0.5\text{T}$ and velocity of ball is $110\text{m}/\text{s}$.

Write the equation for centripetal acceleration.

$a=\frac{v^2}{r}$

Here,

$r$ is the radius of circular path

Rewrite the above relation in terms of $r$.

$r=\frac{v^2}{a}$

Substitute $110\text{m}/\text{s}$ for $v$ and $176\text{m}/{\text{s}}^{\text{2}}$ for $a$ in the above equation to find $r$.

$\begin{array}{c}r=\frac{{\left(110\text{m}/\text{s}\right)}^2}{176\text{m}/{\text{s}}^{\text{2}}}\\ =68.75\text{m}\end{array}$

Conclusion:

Therefore, the radius of circular path is $68.75\text{m}$.