Chapter 19, Problem 122P

(a)

To determine

The direction of the force on the projectile.

Answer to Problem 122P

The force on the projectile is $\text{to the right}$ .

Explanation of Solution

The right hand rule states that in the case of a cross product, if the index finger of the right hand points in the direction of the first vector and the second vector points in the direction of the middle finger, then the cross product will point in the direction of the thumb. Magnetic force on a current carrying wire is the product of the current through the wire and the cross product of the vector, whose magnitude is the length of the wire and is points in the direction of current, and the magnetic field.

Right hand rule for the magnetic force on a current carrying conductor can be described as if the index finger of the right hand points in the direction of current, the middle finger in the direction of the magnetic field, then the thumb will point in the direction of the magnetic force. In the given situation, the current flows into the page and the magnetic field is directed upward. According to the right hand rule, the magnetic force on the projectile is directed to the right.

(b)

To determine

The speed of the projectile after it has travelled $8.00\text{ m}$ down the rails.

Answer to Problem 122P

The speed of the projectile after it has travelled $8.00\text{ m}$ down the rails is $13.6\text{ m/s}$ .

Explanation of Solution

The free body diagram of the projectile is shown in figure 1.

The rod only moves in the $x$ direction so that the net force on the rod in the $y$ direction must be zero.

  $\Sigma F_y=0$                                                                                                                   (I)

Here, $\Sigma F_y$ is the net force on the rod in the $y$ direction.

Refer to figure 1 and write the expression for $\Sigma F_y$ .

  $\Sigma F_y=N-mg$                                                                                                         (II)

Here, $N$ is the normal force, $m$ is the mass of the rod and $g$ is the acceleration due to gravity.

Equate equations (I) and (II).

$\begin{array}{c}N-mg=0\\ N=mg\end{array}$ (III)

Write the equation for the force of kinetic friction.

  $f_k={\mu }_{k}N$

Here, $f_k$ is the magnitude of the force of kinetic friction and ${\mu }_k$ is the coefficient of kinetic friction.

Put equation (III) in the above equation.

  $f_k={\mu }_{k}mg$                                                                                                             (IV)

Apply Newton’s second law to the rod for the motion in $x$ direction.

  $\Sigma F_x=m{a}_x$                                                                                                            (V)

Here, $\Sigma F_x$ is the net force on the rod in the $x$ direction and $a_x$ is the acceleration of the rod.

Refer to figure (I) and write the expression for $\Sigma F_x$ .

  $\Sigma F_x=F_B-f_k$

Here, $F_B$ is the magnitude of the magnetic force on the rod.

Put the above equation in equation (V) and rewrite it for $a_x$ .

  $\begin{array}{c}{F}_B-f_k=m{a}_x\\ a_x=\frac{F_B-f_k}{m}\end{array}$                                                                                                 (VI)

Write the expression for $F_B$ .

  $F_B=ILB$                                                                                                              (VII)

Here, $I$ is the current flowing through the rod, $L$ is the length of the rod and $B$ is the magnitude of the magnetic field.

Put equations (IV) and (VII) in equation (VI).

  $\begin{array}{c}{a}_x=\frac{ILB-{\mu }_{k}mg}{m}\\ =\frac{ILB}{m}-{\mu }_{k}g\end{array}$                                                                                               (VIII)

Write the second law of motion.

  $v_{fx}^2-v_{ix}^2=2a_x\Delta x$

Here, $v_{fx}$ is the final speed of the rod, $v_{ix}$ is the initial speed of the rod and $\Delta x$ is the distance travelled by the rod.

The rod starts from rest so that its initial speed is zero.

Substitute $0$ for $v_{ix}$ in the above equation and rewrite it for $v_{fx}$ .

  $\begin{array}{c}{v}_{fx}^2-0=2a_x\Delta x\\ v_{fx}^2=2a_x\Delta x\\ v_{fx}=\sqrt{2a_x\Delta x}\end{array}$

Put equation (VIII) in the above equation.

  $v_{fx}=\sqrt{2\left(\frac{ILB}{m}-{\mu }_{k}g\right)\Delta x}$                                                                                     (IX)

Conclusion:

Given that the mass of the projectile is $50.0\text{ g}$ , the distance between the rails is $0.500\text{ m}$ , the magnitude of the magnetic field is $0.750\text{ T}$ , the current through the projectile is $2.00\text{ A}$ , the coefficient of kinetic friction is $0.350$ and the distance travelled by the projectile is $8.00\text{ m}$ . The value of the acceleration due to gravity is $9.80{\text{ m/s}}^2$ and the length of the projectile is equal to the distance between the conducting rails.

Substitute $2.00\text{ A}$ for $I$ , $0.500\text{ m}$ for $L$ , $0.750\text{ T}$ for $B$ , $50.0\text{ g}$ for $m$ , $0.350$ for ${\mu }_k$ ,$9.80{\text{ m/s}}^2$ for $g$ and $8.00\text{ m}$ for $\Delta x$ in equation (IX) to find $v_{fx}$ .

  $\begin{array}{c}{v}_{fx}=\sqrt{2\left(\frac{\left(2.00\text{ A}\right)\left(0.500\text{ m}\right)\left(0.750\text{ T}\right)}{\left(50.0\text{ g}\frac{1\text{ kg}}{1000\text{ g}}\right)}-\left(0.350\right)\left(9.80{\text{ m/s}}^2\right)\right)\left(8.00\text{ m}\right)}\\ =13.6\text{ m/s}\end{array}$

Therefore, the speed of the projectile after it has travelled $8.00\text{ m}$ down the rails is $13.6\text{ m/s}$ .