Question

A magnetic field of 37.2 T has been achieved at the MIT Francis Bitter National Magnetic Laboratory. Find the current needed to achieve such a field
(a) 2.00 cm from a long, straight wire;
(b) At the center of a circular coil of radius 42.0 cm that has 100 turns;
(c) Near the center of a solenoid with radius 2.40 cm, length 32.0 cm, and 40,000 turns.

Answer 1

Known:

\(B=37.2 T\)

a) \(2.00 \mathrm{~cm}\) from a long, straight wire

We know that the magnetic field in a straight wire is:

$$ \begin{aligned} &B=\frac{u_{0} I}{2 \pi r} \\ &I=\frac{2 \pi \mathrm{Br}}{u_{0}} \\ &I=\frac{\left(2 \pi \cdot 37.2 T^{*} 2 \times 10^{-2} \mathrm{~m}\right)}{\left(4 \pi \times \times 10^{-7}\right)}=3.72 \times 10^{6} \mathrm{~A} \end{aligned} $$

(c) near the center of a solenoid with radius \(2.40 \mathrm{~cm}\), length \(32.0 \mathrm{~cm}\), and 40,000 turns.

We know that the magnetic field in a solenoid is:

$$ \begin{aligned} &B=\frac{\mu_{0} \mathrm{NI}}{L} \\ &I=\frac{\mathrm{BL}}{\mu_{0} \mathrm{~N}} \\ &I=\frac{\left(37.2 T * 32 \times 10^{-2} \mathrm{~m}\right)}{\left(4 \pi \times 10^{-7} * 40000\right)}=236.823 \mathrm{~A} \end{aligned} $$

Answer 2

Concepts and reason

The given problem can be solved by using the expression of magnetic field due to straight current carrying conductor, at the center of circular coil and near the center of solenoid.

To get the value of current, rearrange the expression of magnetic field due to straight current carrying conductor, at the center of circular coil and near the center of solenoid.

Fundamentals

The expression of magnetic field due to straight current carrying conductor can be expressed as follows,

$B = \frac{{{\mu _0}i}}{{2\pi l}}$

Here, ${\mu _0} = \left( {4\pi \times {{10}^{ - 7}}{\rm{ T}} \cdot {\rm{m}} \cdot {{\rm{A}}^{ - 1}}{\rm{ }}} \right)$ , $i$ is the current, $l$ is the length of the conductor.

The expression of magnetic field at the center of circular of coil can be expressed as follows,

$B = \frac{{{\mu _0}Ni}}{{2r}}$

Here, ${\mu _0} = \left( {4\pi \times {{10}^{ - 7}}{\rm{ T}} \cdot {\rm{m}} \cdot {{\rm{A}}^{ - 1}}{\rm{ }}} \right)$ , $i$ is the current and $r$ is the radius of circular coil.

The expression of magnetic field near the center of the solenoid can be expressed as follows,

$B = \frac{{{\mu _0}Ni}}{l}$

Here, ${\mu _0} = \left( {4\pi \times {{10}^{ - 7}}{\rm{ T}} \cdot {\rm{m}} \cdot {{\rm{A}}^{ - 1}}{\rm{ }}} \right)$ , $i$ is the current and $l$ is the length of the solenoid.

(A)

The expression of magnetic field due to straight current carrying conductor can be expressed as follows,

$B = \frac{{{\mu _0}i}}{{2\pi l}}$

Rearrange the expression to get the value of current.

$i = \frac{{B\left( {2\pi l} \right)}}{{{\mu _0}}}$

Substitute $4\pi \times {10^{ - 7}}{\rm{ T}} \cdot {\rm{m}} \cdot {{\rm{A}}^{ - 1}}$ for ${\mu _0}$ , $37.2{\rm{ T}}$ for $B$ and $2.00{\rm{ cm}}$ for $l$ in the expression of current.

$\begin{array}{c}\\i = \frac{{\left( {37.2{\rm{ T}}} \right)\left( {2\pi \left( {2.00{\rm{ cm}}} \right)\left( {\frac{{1{\rm{ cm}}}}{{100{\rm{ m}}}}} \right)} \right)}}{{4\pi \times {{10}^{ - 7}}{\rm{ T}} \cdot {\rm{m}} \cdot {{\rm{A}}^{ - 1}}}}\\\\ = 3.72 \times {10^6}{\rm{ A}}\\\end{array}$

(B)

The expression of magnetic field at the center of circular of coil can be expressed as follows,

$B = \frac{{{\mu _0}Ni}}{{2r}}$

Rearrange the expression to get the value of current.

$i = \frac{{B\left( {2r} \right)}}{{{\mu _0}N}}$

Substitute $4\pi \times {10^{ - 7}}{\rm{ T}} \cdot {\rm{m}} \cdot {{\rm{A}}^{ - 1}}$ for ${\mu _0}$ , $37.2{\rm{ T}}$ for $B$ , $100$ for $N$ and ${\rm{42}}{\rm{.0 cm}}$ for $r$ in the expression of current.

$\begin{array}{c}\\i = \frac{{\left( {37.2{\rm{ T}}} \right)\left( {2\left( {42.0{\rm{ cm}}} \right)\left( {\frac{{1{\rm{ cm}}}}{{100{\rm{ m}}}}} \right)} \right)}}{{\left( {4\pi \times {{10}^{ - 7}}{\rm{ T}} \cdot {\rm{m}} \cdot {{\rm{A}}^{ - 1}}} \right)\left( {100} \right)}}\\\\ = 2.49 \times {10^5}{\rm{ A}}\\\end{array}$

(C)

The expression of magnetic field near the center of the solenoid can be expressed as follows,

$B = \frac{{{\mu _0}Ni}}{l}$

Rearrange the expression to get the value of current.

$i = \frac{{Bl}}{{{\mu _0}N}}$

Substitute $4\pi \times {10^{ - 7}}{\rm{ T}} \cdot {\rm{m}} \cdot {{\rm{A}}^{ - 1}}$ for ${\mu _0}$ , $37.2{\rm{ T}}$ for $B$ , $40000$ for $N$ and ${\rm{32}}{\rm{.0 cm}}$ for $l$ in the expression of current.

$\begin{array}{c}\\i = \frac{{\left( {37.2{\rm{ T}}} \right)\left( {\left( {32.0{\rm{ cm}}} \right)\left( {\frac{{1{\rm{ cm}}}}{{100{\rm{ m}}}}} \right)} \right)}}{{\left( {4\pi \times {{10}^{ - 7}}{\rm{ T}} \cdot {\rm{m}} \cdot {{\rm{A}}^{ - 1}}} \right)\left( {40000} \right)}}\\\\ = 237{\rm{ A}}\\\end{array}$

Ans: Part A

The value of current is $3.72 \times {10^6}{\rm{ A}}$ .

Part B

The value of current is $2.49 \times {10^5}{\rm{ A}}$ .

Part C

The value of current is ${\rm{237}}\,{\rm{A}}$ .