Chapter 20, Problem 25QAP

To determine

(a)

Magnitude and direction of the force on each side of the coil in problem (24) for situation a), b) and c)

Answer to Problem 25QAP

Each side of b) situation have current of $79.2\text{ mA}$

In top side, force towards bottom

In bottom side, force towards top

In left side, force towards right

In right side, force towards left

In situation a) and b) there is no induced current, so there is no force

Explanation of Solution

Given info:

  $\begin{array}{l}\text{induced current in}\\ \text{situation a) = 0 A}\\ \text{situation b) = 0}\text{.044 A}\\ \text{situation c) = 0 A}\\ \text{Area of the coil= 0}{\text{.01 m}}^{\text{2}}\\ \text{Number of turns = 30}\\ \text{magnetic field = 0}\text{.6 T}\end{array}$

Formula used:

  $\begin{array}{l}F=NIlB\\ F=\text{ magnetic force}\\ B=\text{ magnetic field}\\ l=\text{ side length}\\ \text{I= current}\\ \text{N=number of turns}\end{array}$

Calculation:

There is no induced current in a) and c) so there is no magnetic force in a) and c)

In b), there is an induced current, so the force on sides

  $\text{According to the Lenz's law, current flows counterclockwise to produce a field that goes in to the page}$

  $\begin{array}{l}\text{Top side,}\\ F=NIlB\\ F=30*0.044\text{ A*0}\text{.6 T*0}\text{.1 m}\\ F=79.2\text{ mN}\\ \text{Direction =towards south}\end{array}$

  $\begin{array}{l}\text{Bottom side,}\\ F=NIlB\\ F=30*0.044\text{ A*0}\text{.6 T*0}\text{.1 m}\\ F=79.2\text{ mN}\\ \text{Direction =towards north}\end{array}$

  $\begin{array}{l}\text{left side,}\\ F=NIlB\\ F=30*0.044\text{ A*0}\text{.6 T*0}\text{.1 m}\\ F=79.2\text{ mN}\\ \text{Direction =towards right}\\ \text{right side,}\\ F=NIlB\\ F=30*0.044\text{ A*0}\text{.6 T*0}\text{.1 m}\\ F=79.2\text{ mN}\\ \text{Direction =towards left}\end{array}$

Conclusion:

Each side of b) situation have current of $79.2\text{ mA}$

In top side, force towards bottom

In bottom side, force towards top

In left side, force towards right

In right side, force towards left

In situation a) and b) there is no induced current, so there is no force

To determine

(b)

Magnitude and direction of the force on each side of the coil when it enters from the left

Answer to Problem 25QAP

Each side of have current of $79.2\text{ mA}$

In top side, force towards bottom

In bottom side, force towards top

In left side, force towards right

In right side, force towards left

Explanation of Solution

Given info:

  $\begin{array}{l}\text{induced current in}\\ \text{situation a) = 0 A}\\ \text{situation b) = 0}\text{.044 A}\\ \text{situation c) = 0 A}\\ \text{Area of the coil= 0}{\text{.01 m}}^{\text{2}}\\ \text{Number of turns = 30}\\ \text{magnetic field = 0}\text{.6 T}\end{array}$

Formula used:

  $\begin{array}{l}F=NIlB\\ F=\text{ magnetic force}\\ B=\text{ magnetic field}\\ l=\text{ side length}\\ \text{I= current}\\ \text{N=number of turns}\end{array}$

Calculation:

As the loop enters the field there will be an induced current in the loop

  $\text{According to the Lenz's law, current flows counterclockwise to produce a field that comes out of the page}$

  $\begin{array}{l}\text{Top side,}\\ F=NIlB\\ F=30*0.044\text{ A*0}\text{.6 T*0}\text{.1 m}\\ F=79.2\text{ mN}\\ \text{Direction =towards south}\end{array}$

  $\begin{array}{l}\text{Bottom side,}\\ F=NIlB\\ F=30*0.044\text{ A*0}\text{.6 T*0}\text{.1 m}\\ F=79.2\text{ mN}\\ \text{Direction =towards north}\end{array}$

  $\begin{array}{l}\text{left side,}\\ F=NIlB\\ F=30*0.044\text{ A*0}\text{.6 T*0}\text{.1 m}\\ F=79.2\text{ mN}\\ \text{Direction =towards right}\\ \text{right side,}\\ F=NIlB\\ F=30*0.044\text{ A*0}\text{.6 T*0}\text{.1 m}\\ F=79.2\text{ mN}\\ \text{Direction =towards left}\end{array}$

Conclusion:

Each side have current of $79.2\text{ mA}$

In top side, force towards bottom

In bottom side, force towards top

In left side, force towards right

In right side, force towards left