Question

A thin copper rod 1.00 m long has a mass of 50.0 g. What is the minimum current in the rod that would allow it to levitate above the ground in a magnetic field of magnitude 0.100 T?
a.1.20 A
b.2.40 A
c.4.90 A
d.9.80 A
e.none of these answers

Answer 1

Concepts and reason

The concepts used to solve this problem are magnetic force and gravitational force.

First, equate the magnitude of the magnetic force and the gravitational force acting on the copper rod to determine the condition for levitation.

Use the relationship between the magnetic field acting on the rod, length of the rod, and the acceleration due to gravity to determine the required current in the rod.

Fundamentals

The expression for the magnetic force that acts on the copper rod is as follows:

${F_m} = BIL$

Here, the magnetic force is ${F_m}$, the magnetic field is $B$, the current in the copper wire is $I$, and the length of the wire is $L$.

The expression for the gravitational force is as follows:

${F_g} = mg$

Here, the gravitational force is ${F_g}$, the mass of the copper rod is $m$, the acceleration due to gravity is $g$.

The expression for the condition for levitation of copper rod is as follows:

$\left| {{F_m}} \right| = \left| {{F_g}} \right|$

Here, the magnitude of the magnetic force is $\left| {{F_m}} \right|$ and the magnitude of the gravitational force is $\left| {{F_g}} \right|$

Levitation occurs when the magnetic force on the copper rod is balanced by the gravitational force acting on it.

The expression for this condition is as follows:

$\left| {{F_m}} \right| = \left| {{F_g}} \right|$

Substitute $BIL$ for $\left| {{F_m}} \right|$ and $mg$ for $\left| {{F_g}} \right|$.

$BIL = mg$

Rearrange the above expression for the current.

$I = \frac{{mg}}{{BL}}$

The expression for the current in the copper wire is as follows:

$I = \frac{{mg}}{{BL}}$

Substitute $50.0\,{\rm{g}}$ for $m$, $9.8\,{\rm{m}}{{\rm{s}}^{ - 2}}$ for $g$, $0.100\,{\rm{T}}$ for $B$, and $1.00\,{\rm{m}}$ for $L$.

$\begin{array}{c}\\I = \frac{{\left( {50.0\,{\rm{g}}} \right)\left( {\frac{{1\,{\rm{kg}}}}{{{{10}^3}\,{\rm{g}}}}} \right)\left( {9.8\,{\rm{m}}{{\rm{s}}^2}} \right)}}{{\left( {0.100\,{\rm{T}}} \right)\left( {1.00\,{\rm{m}}} \right)}}\\\\ = \frac{{\left( {50.0 \times {{10}^{ - 3}}\,{\rm{kg}}} \right)\left( {9.8\,{\rm{m}}{{\rm{s}}^2}} \right)}}{{\left( {0.100\,{\rm{T}}} \right)\left( {1.00\,{\rm{m}}} \right)}}\\\\ = 4.90\,{\rm{A}}\\\end{array}$

The required current in the copper wire is $I = 4.90\,{\rm{A}}$.

Ans:

Thus, the correct option for the required current in the copper wire for it to levitate in the magnetic field is $4.90\,A$.