Question

A magnetic field of 37.2 T has been achieved at the MIT Francis Bitter National Magnetic...

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 centre 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 40000 turns

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Answer #1
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=μ0i2πlB = \frac{{{\mu _0}i}}{{2\pi l}}

Here, μ0=(4π×107TmA1){\mu _0} = \left( {4\pi \times {{10}^{ - 7}}{\rm{ T}} \cdot {\rm{m}} \cdot {{\rm{A}}^{ - 1}}{\rm{ }}} \right) , ii is the current, ll is the length of the conductor.

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

B=μ0Ni2rB = \frac{{{\mu _0}Ni}}{{2r}}

Here, μ0=(4π×107TmA1){\mu _0} = \left( {4\pi \times {{10}^{ - 7}}{\rm{ T}} \cdot {\rm{m}} \cdot {{\rm{A}}^{ - 1}}{\rm{ }}} \right) , ii is the current and rr is the radius of circular coil.

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

B=μ0NilB = \frac{{{\mu _0}Ni}}{l}

Here, μ0=(4π×107TmA1){\mu _0} = \left( {4\pi \times {{10}^{ - 7}}{\rm{ T}} \cdot {\rm{m}} \cdot {{\rm{A}}^{ - 1}}{\rm{ }}} \right) , ii is the current and ll is the length of the solenoid.

(A)

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

B=μ0i2πlB = \frac{{{\mu _0}i}}{{2\pi l}}

Rearrange the expression to get the value of current.

i=B(2πl)μ0i = \frac{{B\left( {2\pi l} \right)}}{{{\mu _0}}}

Substitute 4π×107TmA14\pi \times {10^{ - 7}}{\rm{ T}} \cdot {\rm{m}} \cdot {{\rm{A}}^{ - 1}} for μ0{\mu _0} , 37.2T37.2{\rm{ T}} for BB and 2.00cm2.00{\rm{ cm}} for ll in the expression of current.

i=(37.2T)(2π(2.00cm)(1cm100m))4π×107TmA1=3.72×106A\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=μ0Ni2rB = \frac{{{\mu _0}Ni}}{{2r}}

Rearrange the expression to get the value of current.

i=B(2r)μ0Ni = \frac{{B\left( {2r} \right)}}{{{\mu _0}N}}

Substitute 4π×107TmA14\pi \times {10^{ - 7}}{\rm{ T}} \cdot {\rm{m}} \cdot {{\rm{A}}^{ - 1}} for μ0{\mu _0} , 37.2T37.2{\rm{ T}} for BB , 100100 for NN and 42.0cm{\rm{42}}{\rm{.0 cm}} for rr in the expression of current.

i=(37.2T)(2(42.0cm)(1cm100m))(4π×107TmA1)(100)=2.49×105A\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=μ0NilB = \frac{{{\mu _0}Ni}}{l}

Rearrange the expression to get the value of current.

i=Blμ0Ni = \frac{{Bl}}{{{\mu _0}N}}

Substitute 4π×107TmA14\pi \times {10^{ - 7}}{\rm{ T}} \cdot {\rm{m}} \cdot {{\rm{A}}^{ - 1}} for μ0{\mu _0} , 37.2T37.2{\rm{ T}} for BB , 4000040000 for NN and 32.0cm{\rm{32}}{\rm{.0 cm}} for ll in the expression of current.

i=(37.2T)((32.0cm)(1cm100m))(4π×107TmA1)(40000)=237A\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 3.72×106A3.72 \times {10^6}{\rm{ A}} .

Part B

The value of current is 2.49×105A2.49 \times {10^5}{\rm{ A}} .

Part C

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

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