Question

Part A

What current is needed in the solenoid's wires?

Express your answer to two significant figures and include the appropriate units.

Problem 24.16

A researcher would like to perform an experiment in a zero magnetic field, which means that the field of the earth must be canceled. Suppose the experiment is done inside a solenoid of diameter 1.0 m, length 3.6m , with a total of 5000 turns of wire. The solenoid is oriented to produce a field that opposes and exactly cancels the 52 ?T local value of the earth's field.

Problem 24.16 A researcher would like to perform an experiment in a zero magnetic field, which means that the field of the earth must be canceled. Suppose the experiment is done inside a solenoid of diameter 1.0 m, length 3.6 m , with a total of 5000 turns of wire. The solenoid is oriented to produce a field that opposes and exactly cancels the 52 muT local value of the earth's field. Part A What current is needed in the solenoid's wires? Express your answer to two significant figures and include the appropriate units.

Answer 1

Concepts and reason

The concept needed to solve this question is the expression for the magnetic field at center of a solenoid.

Write the expression for the magnetic field at center of a solenoid and solve for current by substituting all the known values.

Fundamentals

The expression for the magnetic field at center of a solenoid is,

$B = {\mu _0}nI$

Here, ${\mu _0}$ is the permeability of free space, n is the number of turns per unit length of the solenoid, and I is the current flow.

The number of turns per unit length of the solenoid is equal to the total number (N) of turns divided by the length (L) of the solenoid.

$n = \frac{N}{L}$

The number of turns per unit length of the solenoid is,

$n = \frac{N}{L}$

Substitute 5000 for N and 3.6 m for L.

$\begin{array}{c}\\n = \frac{{5000}}{{3.6{\rm{ m}}}}\\\\ = {\rm{1388}}{\rm{.9 turns/m}}\\\\ = {\rm{1389 turns/m}}\\\end{array}$

The expression for the magnetic field at center of a solenoid is,

$B = {\mu _0}nI$

Rearrange the equation for I.

$I = \frac{B}{{{\mu _0}n}}$

Substitute 52 ${\rm{\mu T}}$ for B, $4\pi \times {10^{ - 7}}{\rm{ T}} \cdot {\rm{m/A}}$ for ${\mu _0}$ , and ${\rm{1389 turns/m}}$ for n.

$\begin{array}{c}\\I = \frac{{52{\rm{ \mu T}}\left( {\frac{{{{10}^{ - 6}}{\rm{ T}}}}{{1{\rm{ \mu T}}}}} \right)}}{{\left( {4\pi \times {{10}^{ - 7}}{\rm{ T}} \cdot {\rm{m/A}}} \right)\left( {{\rm{1389 turns/m}}} \right)}}\\\\ = {\rm{2}}{\rm{.98}} \times {10^{ - 2}}{\rm{ A}}\\\\ = 0.0298{\rm{ A}}\\\\ = 0.03{\rm{ A}}\\\end{array}$

Ans:

The required value of current is $0.03{\rm{ A}}$ .