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Compare the magnitude of the magnetic field at the center of a circular current loop of radius 20 mm with the magnitude of the magnetic field at the center of a solenoid of the same radius and with 2.0 turn per millimeter. Assume the current is the same through the current loop and the solenoid.

Answer :

Answer:

The magnitude of the magnetic field at the center of a solenoid is 80 times of the magnetic field at the center of a loop.

Explanation:

Given that,

Radius = 20 mm

Number of turns = 2000

Suppose, we can use wire of 1.0 mm diameter to make either a single loop of wire or a solenoid, and we wish to compare, for a given current, the magnetic fields at the center of each.

Suppose a small element of wire of length dl.

Let I be the current flowing through the wire

We need to calculate the magnetic field

Using formula of magnetic field

[tex]B=\int{\dfrac{\mu_{0}}{4\pi}\times\dfrac{I\times dl}{R^2}}[/tex]

[tex]B=\dfrac{\mu_{0}I}{4\pi\times R^2}\int{dl}[/tex]

Put the value into the formula

[tex]B=\dfrac{\mu_{0}I}{4\pi\times r^2}\times2\pi R[/tex]

[tex]B=\dfrac{\mu_{0}I}{2R}[/tex]...(I)

We need to calculate the magnetic field due to solenoid

Using formula of magnetic field

[tex]B'=\mu_{0}nI[/tex]...(II)

We need to calculate the magnitude of the magnetic field at the center of a circular current loop

Dividing equation (II) by equation (I)

[tex]\dfrac{B'}{B}=\dfrac{\mu_{0}nI}{\dfrac{\mu_{0}I}{2R}}[/tex]

[tex]\dfrac{B'}{B}=2nR[/tex]

Put the value into the formula

[tex]\dfrac{B'}{B}=2\times2000\times20\times10^{-3}[/tex]

[tex]\dfrac{B'}{B}=80[/tex]

[tex]B'=80 B[/tex]

Hence, The magnitude of the magnetic field at the center of a solenoid is 80 times of the magnetic field at the center of a loop.

The magnitude of the magnetic field at the center of a solenoid is [tex]80[/tex]times of the magnetic field at the center of a loop.

Magnetic field strength :

The magnitude of the magnetic field at the center of a circular current loop is given as,

                    [tex]B_{L}=\frac{\mu _{0}I}{2R}[/tex]

the magnetic field at the center of a solenoid is,

                    [tex]B_{S}=\mu_{0}nI[/tex]

Where

  • [tex]\mu_{0}[/tex] magnetic permeability
  • [tex]I[/tex] is current
  • [tex]R[/tex] is radius.

Taking the ratio of both

                  [tex]\frac{B_{S}}{B_{L}}=2nR[/tex]

Given that, [tex]n=2000,R=20mm=20*10^{-3}m[/tex]

              [tex]\frac{B_{S}}{B_{L}}=2*2000*20*10^{-3} \\\\\frac{B_{S}}{B_{L}}=80\\\\B_{s}=80B_{L}[/tex]

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