Chapter 14, Problem 2SP

(a)

To determine

The magnitude of magnetic force on the ball.

Answer to Problem 2SP

Magnetic force on the ball is $7.2\text{N}$.

Explanation of Solution

Given Info The charge of metal ball is $+0.09\text{C}$, magnetic field is $0.4\text{T}$ and velocity of ball is $200\text{m}/\text{s}$.

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 $+0.09\text{C}$ for $q$,$200\text{m}/\text{s}$ for $v$ and $0.4\text{T}$ for $B$ in the above equation to find $F$.

$\begin{array}{c}F=\left(+0.09\text{C}\right)\left(200\text{m}/\text{s}\right)\left(0.4\text{T}\right)\\ =7.2\text{N}\end{array}$

Conclusion:

Therefore, the force is $7.2\text{N}$.

(b)

To determine

The direction of magnetic force on the ball.

Answer to Problem 2SP

Magnetic force upwards.

Explanation of Solution

Direction of magnetic force on ball is given by the right hand thumb rule. The rule says that 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.

In the figure, the magnetic field acts into the page and is perpendicular to the page. The ball moves perpendicular to the magnetic field to the right. Pointing the index finger of the right hand in the direction of motion of ball and middle finger into the page makes the thumb to point upward.

Conclusion:

Therefore, the Magnetic force acts upwards.

(c)

To determine

Whether the force on ball will change the magnitude of the velocity of ball.

Answer to Problem 2SP

The force on ball will not change the magnitude of the velocity of ball, it can only change the direction of velocity.

Explanation of Solution

Velocity is a vector quantity having both magnitude and direction. The magnetic force acting on the ball is perpendicular to the direction of motion of the ball. This makes the ball to have a circular motion. The direction of velocity in circular motion is changing at every instant since the direction is along the tangent at a particular instant.

The magnitude of the direction of the velocity is given as the ratio of the magnetic force to the product of the charge of the ball and the magnitude of the magnetic field. All the quantities in the ratio are constant so that the ratio will be constant. Since the ratio is a constant the magnitude of the velocity also will be a constant.

Conclusion:

Therefore, the force on ball will not change the magnitude of the velocity of ball.

(d)

To determine

The magnitude of the acceleration of the charged ball.

Answer to Problem 2SP

The magnitude of the acceleration of the charged ball is $240\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Given Info The mass of the ball is $0.030\text{kg}$.

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 $7.2\text{N}$ for $F$ and $0.030\text{kg}$ for $m$ in the above equation to find $a$.

 $\begin{array}{c}a=\frac{7.2\text{N}}{0.030\text{kg}}\\ =240\text{m}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the acceleration of the charged ball is $240\text{m}/{\text{s}}^{\text{2}}$.

(e)

To determine

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

Answer to Problem 2SP

The radius of circular path is $167\text{m}$,

Explanation of Solution

Given Info The velocity of the ball is $200\text{m}/\text{s}$.

Write the equation for centripetal acceleration.

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

Here,

$r$ is the radius of circular path

Rewrite the above relation for $r$.

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

Substitute $200\text{m}/\text{s}$ for $v$ and $240\text{m}/{\text{s}}^{\text{2}}$ for $a$ in the above equation to find $r$.

$\begin{array}{c}r=\frac{{\left(200\text{m}/\text{s}\right)}^2}{240\text{m}/{\text{s}}^{\text{2}}}\\ =167\text{m}\end{array}$

Conclusion:

Therefore, the radius of circular path is. $167\text{m}$