Solved Problem on Magnetic Force

The horizontal wire in the figure has a mass of 50 g, a length of 2.0 m, and rises with an unknown acceleration. We know that in the region there is a horizontal magnetic field of 4×10⁻¹ tesla, perpendicular to the wire and that the wires are traversed by a current of 1.5 amperes.
a) Determine the direction of the field;
b) Calculate the magnetic force;
c) Calculate the acceleration.

Problem data:

Mass of the wire: m = 50 g;
Wire length: L = 2.0 m;
Magnetic field: B = 4×10⁻¹ T;
Electric current: i = 1.5 A.

Solution

Firstly, we convert the unit of mass given in grams (g) to kilograms (kg) used in the International System of Units (SI).

\[ \begin{gather} m=50\;\cancel{\text{g}}\times\frac{10^{-3}\;\text{kg}}{1\;\cancel{\text{g}}}=50\times10^{-3}\;\text{kg}=5\times10^{-2}\;\text{kg}=0.05\;\text{kg} \end{gather} \]

a) To determine the direction of the magnetic field, we use the left-hand rule, placing the thumb in the direction of the acceleration of the wire, we have the direction of the force (by Newton's Second Law, \( \vec{F}=m\vec{a} \), the force and the acceleration have the same direction) the middle finger is placed in the direction of the current, then the index finger will give the direction of the field (Figure 1-A)

Figure 1

In Figure 1-B, we have the elements of the problem seen in perspective, and in Figure 1-C, using the diagram of the problem, we have that the electric field is pointing into the sheet.

b) The magnitude of the magnetic force, F_M, is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {F_{M}=BiL} \end{gather} \]

\[ \begin{gather} F_{M}=4\times 10^{-1}\times 1.5\times 2.0 \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {F_{M}=1.2\;\text{N}} \end{gather} \]

c) Two forces act on the wire, the magnetic force, \( {\vec{F}}_{M} \), calculated in the previous item, and the gravitational force, \( \vec{P} \). To calculate the acceleration of the wire, we use Newton's Second Law

\[ \begin{gather} \bbox[#99CCFF,10px] {\vec{F}=m\vec{a}} \tag{I} \end{gather} \]

the resultant \( \vec{F} \) in the problem is given by the difference between the magnetic force and the gravitational force (Figure 2).

Figure 2

The gravitational force is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {F_{g}=mg} \tag{II} \end{gather} \]

Then applying expression (I) to the vertical motion, we have the acceleration

\[ \begin{gather} F_{M}-F_{g}=ma \tag{III} \end{gather} \]

substituting expression (II) into expression (III)

\[ \begin{gather} F_{M}-mg=ma\\ a=\frac{F_{M}-mg}{m} \end{gather} \]

substituting the numerical values

\[ a=\frac{1.2-0.05\times 9.8}{0.05} \]

\[ \bbox[#FFCCCC,10px] {a=14.2\;\text{m/s}^{2}} \]

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .