Chapter 20, Problem 75GP

(a)

To determine

To Find: The speed of rod as a function of time assuming no friction between rail and rod.

Answer to Problem 75GP

  $v=\frac{IlB}{m}t$

Explanation of Solution

Given info:

Distance between the wire $=l$

Mass of light rod $=m$

Magnetic field $=B$

Current through the system $=I$

Formula used:

Magnetic force:

  $F=IlB\sin \theta$

Where,

  $\theta =$ Angle between current element and magnetic field.

Newton’s second law:

  $F=m\frac{d^{2}x}{d{t}^2}$

Where,

  $\frac{d^{2}x}{d{t}^2}=$ Acceleration.

Calculation:

Magnetic force acting on rod would be equal to Newton’s second law:

i.e.

  $\begin{array}{c}m\frac{d^{2}x}{d{t}^2}=IlB\sin {90}^o\\ \frac{d^{2}x}{d{t}^2}=\frac{IlB}{m}\\ \frac{d}{dt}\left(\frac{dx}{dt}\right)=\frac{IlB}{m}\\ \frac{d}{dt}\left(v\right)=\frac{IlB}{m}\\ dv=\frac{IlB}{m}dt\end{array}$

Integrating both side:

  $\begin{array}{c}{\displaystyle {\int }_0^{v}dv}={\displaystyle {\int }_0^t\frac{IlB}{m}dt}\\ v=\frac{IlB}{m}t\end{array}$

Conclusion:

Speed of rod as a function of time: $v=\frac{IlB}{m}t$

(b)

To determine

To Find: The speed of rod as a function of time in the presence of friction between rail and rod.

Answer to Problem 75GP

  $v=\left(\frac{IlB}{m}-{\mu }_{k}g\right)t$

Explanation of Solution

Given info:

Distance between the wire $=l$

Mass of light rod $=m$

Magnetic field $=B$

Current through the system $=I$

Formula used:

Magnetic force:

  $F=IlB\sin \theta$

Where,

  $\theta =$ Angle between current element and magnetic field.

Newton’s second law:

  $F=m\frac{d^{2}x}{d{t}^2}$

Where,

  $\frac{d^{2}x}{d{t}^2}=$ Acceleration.

Friction force:

  $f_k={\mu }_{k}mg$

Calculation:

From Newton’s second law,

  $\begin{array}{c}m\frac{d^{2}x}{d{t}^2}=F_{net}\\ m\frac{d^{2}x}{d{t}^2}=IlB\sin {90}^o-{\mu }_{k}mg\\ \frac{d^{2}x}{d{t}^2}=\frac{IlB}{m}-{\mu }_{k}g\\ \frac{d}{dt}\left(\frac{dx}{dt}\right)=\frac{IlB}{m}-{\mu }_{k}g\\ \frac{d}{dt}\left(v\right)=\frac{IlB}{m}-{\mu }_{k}g\\ dv=\left(\frac{IlB}{m}-{\mu }_{k}g\right)dt\end{array}$

Integrating both side:

  $\begin{array}{c}{\displaystyle {\int }_0^{v}dv}={\displaystyle {\int }_0^t\left(\frac{IlB}{m}-{\mu }_{k}g\right)dt}\\ v=\left(\frac{IlB}{m}-{\mu }_{k}g\right)t\end{array}$

Conclusion:

Speed of rod as a function of time in the presence of friction: $v=\left(\frac{IlB}{m}-{\mu }_{k}g\right)t$

(c)

To determine

To Find: The moving direction of rod if the current through it heads north.

Explanation of Solution

Introduction:

Fleming’s left-hand rule:

When a current carrying conductor is placed in a magnetic field, the direction of motion of the conductor can be given by Fleming’s left-hand rule.

In Fleming’s left-hand rule, middle finger is directed towards current and fore finger directed towards the magnetic field and therefore, the direction of thumb gives the direction of motion of conductor.

Pointing the middle finger of left hand towards north and fore finger out of the plan, the thumb is directed towards the east.

Conclusion:

Hence, the direction of motion of the rod is towards the east.