Chapter 22, Problem 73PE

To determine

(a)

The magnitude and direction of magnetic force on the bob at the lowest point of the pendulum when it is released.

Answer to Problem 73PE

  $9.0\times {10}^{-7}N$ in upward direction.

Explanation of Solution

Given:

A pendulum is set up so that its bob swings between the poles of permanent magnet. Charge on the bob is given as $0.25\mu C=0.25\times {10}^{-6}C$ and height from bottom of the bob where it is released is $30.0cm$. The uniform magnetic field in which the bob is swinging is $1.50T$.

Formula used:

The magnetic force on a charge particle having charge $q$ moving in a uniform magnetic field $B$ with velocity $v$ is $\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$. If the velocity vector and the magnetic field vector is perpendicular then the force is $F=qvB$.

Calculation:

A swinging bob of a pendulum is released from rest from height $30.0cm$. Let mass of the bob is $m$. After releasing the potential energy of the bob is converted into kinetic energy and the bob will gain a tangential velocity.

  $\begin{array}{l}PE=KE\\ mgh=\frac{m{v}^2}{2}\end{array}$

The tangential velocity is then $v=\sqrt{2gh}$ .----(1)

Substituting the values of acceleration due to gravity $g$ and height from where the bob is released. We get

  $v=\sqrt{2gh}=\sqrt{2\times 9.8\times 0.3}=2.4m/s$

Now the force on the bob is $F=qvB$ .----(2)

Substituting the values of charge on the bob $q$, tangential velocity $v$, and the magnetic field $B$. We get

  $F=qvB=(0.25\times {10}^{-6})(2.4)(1.5)=0.9\times {10}^{-6}=9.0\times {10}^{-7}N$

And the direction of the force is upward.

Conclusion:

When the bob of pendulum placed in the magnetic field region is released, the magnetic force on itis $9.0\times {10}^{-7}N$ in upward direction.

To determine

(b)

The acceleration of the bob at the bottom of its swing in a magnetic field region.

Answer to Problem 73PE

  $3.0\times {10}^{-5}m/s^2$

Explanation of Solution

Given:

A pendulum is set up so that its bob swings between the poles of permanent magnet. Charge on the bob is given as $0.25\mu C=0.25\times {10}^{-6}C$ and height from bottom of the bob where it is released is $30.0cm$. The uniform magnetic field in which the bob is swinging is $1.50T$. Mass of the bob is $30.0g=30\times {10}^{-3}kg$.

Formula used:

The magnetic force on a charge particle having charge $q$ moving in a uniform magnetic field $B$ with velocity $v$ is $\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$. If the velocity vector and the magnetic field vector is perpendicular then the force is $F=qvB$.

Force on a particle of mass $m$ accelerating with acceleration $a$ is $F=ma$.

Calculation:

It is seen in part (a) that the magnetic force acting on the bob when it is released from height in a uniform magnetic field is $9.0\times {10}^{-7}N$. As the bob is released from a height it gains some kinetic energy that will accelerates the bob. So the magnetic force must be equal to mass times acceleration produced in the bob.

  $F=ma$

  $9.0\times {10}^{-7}$

  $=ma$

The acceleration produced in the bob is $a=\frac{9.0\times {10}^{-7}}{m}$. Substituting the value of mass of the bob. We get acceleration of the bob is $a=\frac{9.0\times {10}^{-7}}{30\times {10}^{-3}}=3.0\times {10}^{-5}m/s^2$

Conclusion:

The acceleration of the bob at the bottom of its swing is $3.0\times {10}^{-5}m/s^2$.