Question

A 2.60-N metal bar, 0.850m long and having a resistance of 10.0? , rests horizontally on conducting wires connecting it to the circuit shown in (Figure 1) . The bar is in a uniform, horizontal, 1.60-T magnetic field and is not attached to the wires in the circuit.

What is the acceleration of the bar just after the switchS is closed?

Answer 1

Concepts and reason

The concepts use to solve this problem are current division rule and force on a current carrying conductor in presence of a magnetic field.

First calculate the equivalent resistance of the circuit by using series and parallel combination of the circuit. After that calculate the total current in the circuit by using Ohm’s law. Next calculate the magnetic force experience on the metal bar. Finally calculate the acceleration of the bar just after the switch $S$ is closed.

Fundamentals

Current division rule, if current flow through more than one parallel paths, each of the parallel path shares a portion of the current depending on the impedance of the path.

A current carrying conductor when placed in presence of a magnetic field experiences a magnetic force. The direction of magnetic force is given by right hand rule where when all the fingers are pointed in the direction of magnetic field with thumb pointing in direction of current, the direction perpendicular to palm of right hand gives the direction of magnetic force.

From ohm’s law,

$V = IR$

Here, $V$ is voltage, $I$ is current, and $R$ is resistance.

The expression of magnetic force acting on the current carrying conductor, as placed in magnetic field,

${F_{\rm{m}}} = BIL\sin \theta$

Here, ${F_{\rm{m}}}$ is magnetic force, $B$ is magnetic field, $I$ is current, $L$ is length of conductor and $\theta$ is angle between current and magnetic field.

The expression of force in term of mass and acceleration.

$F = ma$

Here, $m$ is mass, and $a$ is acceleration.

The expression for the current in an arm when flowing through two parallel arms is given by,

Calculate the equivalent resistance of the circuit just after the switch is closed.

As the switch is closed, the metal bar and the $1{\rm{0 \Omega }}$ resistor forms a parallel combination of resistors. The equivalent resistance of this parallel combination is,

$\begin{array}{c}\\{{\rm{Z}}_{\rm{P}}}{\rm{ = }}\left( {\frac{{{\rm{10 \times 10}}}}{{{\rm{10 + 10}}}}} \right)\Omega \\\\ = {\rm{5 \Omega }}\\\end{array}$

Here, ${{\rm{Z}}_{\rm{P}}}$ is equivalent resistance in parallel combination

Now calculate the equivalent resistance of the circuit.

The $25{\rm{ }}\Omega$ resistor and ${Z_{\rm{p}}}$ are in series so,

$\begin{array}{c}\\{{\rm{Z}}_{{\rm{eq}}}} = {\rm{25 }}\Omega + {\rm{5 \Omega }}\\\\ = 3{\rm{0 \Omega }}\\\end{array}$

Now calculate the total current in the circuit.

From ohm’s law,

$V = IR$

Here, $V$ is voltage, $I$ is current, and $R$ is resistance.

Rearrange the equation $V = IR$ for current $I$ .

$I = \frac{V}{R}$

Substitute $120.0{\rm{ V}}$ for $V$ and $3{\rm{0 \Omega }}$ for $R$ .

$\begin{array}{c}\\I = \frac{{120.0{\rm{ V}}}}{{3{\rm{0 \Omega }}}}\\\\ = 4.0{\rm{ A}}\\\end{array}$

Calculate the current in the metal bar by using current division rule.

From the current division rule,

${I_{\rm{m}}} = \frac{{{R_2}}}{{{R_1} + {R_2}}}I$

Here, ${{\rm{I}}_{\rm{m}}}$ is current through metal bar, $I$ is total current through the circuit and ${R_1},{R_2}$ are the resistances in two parallel paths.

Substitute $4{\rm{ A}}$ for $I$ , $1{\rm{0 \Omega }}$ for ${{\rm{R}}_{\rm{2}}}$ and $1{\rm{0 \Omega }}$ for ${{\rm{R}}_1}$ .

$\begin{array}{c}\\{I_{\rm{m}}}{\rm{ = }}\left( {\frac{{{\rm{10}}}}{{{\rm{10 + 10}}}}} \right){\rm{4}}\\\\ = {\rm{2 A}}\\\end{array}$

Calculate the force experience on the metal bar.

The expression of magnetic force acting on the current carrying conductor, as placed in magnetic field,

${F_{\rm{m}}} = BIL\sin \theta$

Here, ${F_m}$ is magnetic force, $B$ is magnetic field, $I$ is current, $L$ is length of conductor and

$\theta$ is angle between current and magnetic field.

Substitute $1.60{\rm{ T}}$ for, ${\rm{2 A}}$ for ${\rm{I}}$ , $90^\circ$ for $\theta$ , and $0.8{\rm{5 m}}$ for $L$ .

$\begin{array}{c}\\{F_{\rm{m}}} = \left( {1.60{\rm{ T}}} \right)\left( {{\rm{2 A}}} \right)\left( {0.8{\rm{5 m}}} \right)\left( {\sin 90^\circ } \right)\\\\ = \left( {1.60{\rm{ T}}} \right)\left( {{\rm{2 A}}} \right)\left( {0.8{\rm{5 m}}} \right)\left( 1 \right)\\\\ = 2.7{\rm{2 N}}\\\end{array}$

Calculate the mass of the metal bar.

The expression of weight is,

$w = mg$

Here, $w$ is weight, $m$ is mass, and $g$ is acceleration due to gravity.

Rearrange the equation $w = mg$ for $m$ .

$m = \frac{w}{g}$

Substitute ${\rm{2}}{\rm{.60 N}}$ for $w$ , and ${\rm{9}}{\rm{.8 m/}}{{\rm{s}}^2}$ for $g$ .

$\begin{array}{c}\\m = \frac{{{\rm{2}}{\rm{.60 N}}}}{{{\rm{9}}{\rm{.8 m/}}{{\rm{s}}^2}}}\\\\ = 0.265{\rm{ kg}}\\\end{array}$

Now calculate the acceleration of the metal bar.

The expression of force in term of mass and acceleration.

$F = ma$

Here, $m$ is mass, and $a$ is acceleration.

Rearrange the equation $F = ma$ for $a$ .

$a = \frac{F}{m}$

Substitute $2.7{\rm{2 N}}$ for ${F_{\rm{m}}}$ and $0.265{\rm{ kg}}$ for $m$ .

$\begin{array}{c}\\a = \frac{{2.7{\rm{2 N}}}}{{0.265{\rm{ kg}}}}\\\\ = 10.26{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

Ans:

The acceleration of the metal bar is $10.26{\rm{ m/}}{{\rm{s}}^2}$ .