Chapter 20, Problem 7P

(a)

To determine

The type of magnetic pole is the top pole face.

Answer to Problem 7P

The top face of the magnetic pole is South Pole.

Explanation of Solution

Introduction:

According to the right-hand rule, if direction of current flowing is indicated by thumb then the curled fingers will represents the direction of magnetic field.

The current flows horizontally towards the right direction and the magnetic field is vertical. Thus, according to the right-hand rule, the top pole face must be the South Pole.

Conclusion:

The top face of the magnetic pole is the South Pole.

(b)

To determine

The current in the wire.

Answer to Problem 7P

The current in the wire is 4.1A.

Explanation of Solution

Given:

When placed in between the pole faces of a magnet, a wire carries the maximum force of $6.50\times {10}^{-2} \text{N}$ . The diameter of the pole face is 10 cm when the magnetic field applied is 0.16T.

Formula Used:

The force applied in the wire due to the magnetic field is

  $\begin{array}{l}F=I(l\times B)=IlBsin\theta \\ F_{\max }=IlB\end{array}$

Calculation:

Since the wire is in between the magnets. So, the length of the wire will be equal to the diameter of the pole face. Thus,

  $\begin{array}{c}{F}_{\max }=IlB\\ I=\frac{F_{\max }}{lB}=\frac{\text{6}{\text{.5×10}}^{\text{-2}}\text{N}}{\text{(0}\text{.1m)(0}\text{.16T)}}\\ =4.063\text{A}\\ I\approx 4.1\text{A}\end{array}$

Conclusion:

The current in the wire is $4.1 \text{A}$

(b)

To determine

New force experienced.

Answer to Problem 7P

The force experienced by the wire is $6.40\times {10}^{-2} \text{N}$

Explanation of Solution

Given:

The diameter of the pole face is 10 cm when the magnetic field applied is 0.16T. The wire makes an angle of $10°$ with the vertical.

Formula Used:

The force applied in the wire due to the magnetic field is

  $\begin{array}{l}F=I(l\times B)=IlBsin\theta \\ F_{\max }=IlB\end{array}$

Calculation:

The angle measured to the horizontal is $90°-10°=80°$

Thus, the force experienced by the tipped wire is

  $\begin{array}{c}F=I(l\times B)\\ =IlBsin\theta \\ F=(6.5\times {10}^{-2}\text{N})sin80°\\ =6.4\times {10}^{-2}\text{N}\end{array}$

Conclusion:

The force experienced by the wire is $6.40\times {10}^{-2} \text{N}$