Question

A 1.0 kg copper rod rests on two horizontal rails 1.0 m apart and carries a current of 50A from one rail to the other. The coefficient of static friction between rod and rails is 0.60. (a) What are the magnitude and (b) the angle (relative to the vertical) of the smallest magnetic field that puts the rod on the verge of sliding?

Short answer

1. The magnitude of the smallest magnetic field is 0.10 T.
2. The angle (relative to the vertical) of the smallest magnetic field is 31^o.

Step-by-step solution

Write the given data

1. The mass of the copper rod, m = 1.0kg
2. The distance between the horizontal rails, d = 1.0m
3. The current flowing from one rail to the other, i = 50A
4. The coefficient of static friction between the rod and rail,${\mu }_s=0.60$

Determine the formulas of magnetic force

The magnetic force along a loop wire is as follows:

$\stackrel{\rightharpoonup }{d{F}_B}=i\left(\stackrel{\rightharpoonup }{dL}\times \stackrel{\rightharpoonup }{B}\right)$ ….. (i)

Here, i is the current in the loop, $\stackrel{\rightharpoonup }{B}$is the magnetic field vector that it experiences, $\stackrel{\rightharpoonup }{dL}$is the length vector of the conducting wire.

The force according to Newton’s second law is as follows:

F = ma ….. (ii)

Here, m is the mass of the body, a is the acceleration of the body in motion.

The static frictional force between two bodies is as follows:

$f={\mu }_{s}N$ …… (iii)

Here, ${\mu }_s$is the static coefficient of static friction, Nis the normal force acting on the body.

a) Determine the smallest magnetic field

The horizontal magnetic force acts on the rod to overcome the force of friction. Due to its orientations, it reduces both normal force and the force of friction. The forces acting on the rod are magnetic force $\stackrel{\rightharpoonup }{F}$, the gravitational force mg, the normal force acting on the rod in the upward direction N and the force of friction f.

The expression for the frictional force acting on the rod is given using equation (iii) as:

$f_s={\mu }_{s}N$

Consider that the rod is moving in the east direction. Then the frictional force acting on the rod will move in the opposite, that is,in the west direction. Suppose F_x is the horizontal eastward component and F_y is the upward component of the magnetic force. Consider the current flowing in the rod in the northward direction.

By using the right hand rule, a downward component B_d of $\stackrel{\rightharpoonup }{B}$ will produce an eastward and a westward component B_w will produce an upward F_y .

Thus, the horizontal and vertical components of the magnetic force can be given using equation (i) as follows:

$$\begin{gathered} F_x=iL{B}_d \\ {F}_y=iL{B}_w \end{gathered}$$

Consider the forces acting along vertical axes. By using Newton’s second law, the force equation can be given as follows:

$\begin{array}{rcl}N& =& mg-F_y\\ & =& mg-iL{B}_w\end{array}$

Thus, the frictional force acting on the body in horizontal direction can be given using the above value in equation (iii) as follows:

$\begin{array}{rcl}f& =& f_{smax}\\ & =& {\mu }_s\left(mg-iL{B}_w\right)\end{array}$

It is on the verge of motion; hence the horizontal acceleration is zero.

Thus, the net horizontal component of the force is zero, that will given a magnetic field equation as follows:

role="math" localid="1662959944352" width="322" height="112">fx-f=0iLBd-f=0iLBd=fiLBd=μsmg-iLBw………………………..(iv)

The horizontal and vertical components of the magnetic field can be given as:

$$\begin{gathered} B_w=B\sin \theta \\ {B}_d=B\cos \theta \end{gathered}$$

Thus, using equation (iv), the magnetic field can be given using the above values as follows:

role="math" localid="1662959969752" $\begin{array}{rcl}iLB\cos \theta & =& {\mu }_s\left(mg-iLB\sin \theta \right)\\ iLB\cos \theta & =& {\mu }_{s}mg-{\mu }_{s}iLB\sin \theta \\ iLB\cos \theta +{\mu }_{s}iLB\sin \theta & =& {\mu }_{s}mg\\ B& =& \frac{{\mu }_{s}mg}{iL\left(\cos \theta +{\mu }_s\sin \theta \right)}\dots \dots \dots \dots \dots \dots \dots \dots \dots ..\left(v\right)\end{array}$

To find the angle, we can differentiate equation (v) with respect to $\theta$, as follows:

role="math" localid="1662960392845" $\begin{array}{rcl}\frac{dB}{d\theta }& =& \frac{{\mu }_{s}mg}{iL}{\left(\cos \theta +{\mu }_s\sin \theta \right)}^{-1}\\ & =& \frac{{\mu }_{s}mg}{iL}\left(\cos \theta +{\mu }_s\sin \theta \right)\\ & =& \frac{{\mu }_{s}mg}{iL}\left(-\sin \theta -{\mu }_s\cos \theta \right)\dots \dots \dots \dots \dots \dots \dots \dots \dots ..\left(vi\right)\end{array}$

But, the value of the magnetic field with the angle needs to be equated to zero value to get the angle giving the smallest magnetic field as follows:

$\begin{array}{rcl}\frac{dB}{d\theta }& =& 0\\ \frac{{\mu }_{s}mg}{iL}\left(-\sin \theta -{\mu }_s\cos \theta \right)& =& 0\\ \frac{{\mu }_{s}mg}{iL}\sin \theta & =& \frac{{{\mu }_s}^{2}mg}{iL}\cos \theta \\ \tan \theta & =& {\mu }_s\end{array}$

Resolve further as:

$$\begin{gathered} \theta ={\tan }^{-1}\left({\mu }_s\right) \\ \theta ={\tan }^{-1}\left(0.60\right) \\ \theta ={31}^o \end{gathered}$$

Thus, the minimum magnetic field is given using the above value in equation (v) as follows:

$\begin{array}{rcl}{B}_{min}& =& \frac{\left(0.60\right)\left(1.0kg\right)\left(9.8{\displaystyle \frac{m}{s^2}}\right)}{\left(50A\right)\left(1m\right)\left(\cos \left({31}^o\right)+0.6\sin \left({31}^o\right)\right)}\\ & =& 0.10T\end{array}$

Hence, the value of the smallest magnetic field is 0.10 T.

b) Calculation of the angle related to the smallest field value

The angle (relative to the vertical) of the smallest magnetic field is given by the calculations in part (a) as:$\theta ={31}^o$

Hence, the value of the required angle is 31^o.