Question

A)Using Biot-Savart Law,obtain an expression for the magnetic field along the axis of a current loop. B)In the Bohr model of hydrogen atom, the electron follows a circular orbit centred on the nucleus containing a proton. The motion of the electron along the circular orbit constitutes a current. Calculate the magnetic field produced by the orbiting electron at the site of the proton. C)A fuse in an electric circuit is designed to open the circuit like a switch when the current exceeds a preset value. If a fuse is made of material that melts when the current density reaches 400 A cm^2, what is the diameter of the wire needed to limit the current to 0.30 A?

Answer 1

Answer on Question #50821-Physics-Field Theory

A) Using Biot-Savart Law, obtain an expression for the magnetic field along the axis of a current loop.

B) In the Bohr model of hydrogen atom, the electron follows a circular orbit centered on the nucleus containing a proton. The motion of the electron along the circular orbit constitutes a current. Calculate the magnetic field produced by the orbiting electron at the site of the proton.

C) A fuse in an electric circuit is designed to open the circuit like a switch when the current exceeds a preset value. If a fuse is made of material that melts when the current density reaches $j = 400\frac{A}{cm^2}$ , what is the diameter of the wire $d$ needed to limit the current to $I_{lim} = 0.30A$ ?

Solution

A) $dB = \frac{\mu_0}{4\pi}\frac{IdL}{r^2}$

$$r ^ {2} = z ^ {2} + R ^ {2}.$$

Substituting $dB_{z} = dB\sin \theta$ and $\sin \theta = \frac{R}{\sqrt{z^2 + R^2}}$ gives

$$d B _ {z} = \frac {\mu_ {0} I d L}{4 \pi} \frac {R}{\left(z ^ {2} + R ^ {2}\right) ^ {\frac {3}{2}}}.$$

The application of the Biot-Savart law on the centerline of a current loop involves integrating the $z$ -component.

The symmetry is such that all the terms in this element are constant except the distance element $dL$ , which when integrated just gives the circumference of the circle. The magnetic field is then

$$B _ {z} = \frac {\mu_ {0}}{4 \pi} \frac {2 \pi R ^ {2} I}{(z ^ {2} + R ^ {2}) ^ {\frac {3}{2}}}.$$

B) An expression for the magnetic field at the center of a current loop is

$$B = \frac {\mu_ {0}}{2} \frac {I}{R},$$

where $R$ is a raius of orbit. A current is

$$I = \frac {e}{T} = e f.$$

So,

$$B = \frac {\mu_ {0}}{2} \frac {e f}{R}.$$

Frequency of electron is

$$f _ {n} = \frac {n \hbar}{m R _ {n}}.$$

Thus

$$B = \frac {\mu_ {0}}{2} \frac {e n \hbar}{m R _ {n} ^ {2}}.$$

For $n = 1$:

$$B = \frac {\mu_ {0}}{2} \frac {\left(\frac {e}{m}\right) \hbar}{a _ {0} ^ {2}}.$$

$a_0$ is Bohr radius.

$$B = \frac {4 \pi \cdot 1 0 ^ {- 7}}{2} \frac {1 . 7 5 9 \cdot 1 0 ^ {1 1} \cdot 1 . 0 5 4 \cdot 1 0 ^ {- 3 4}}{(0 . 0 5 9 \cdot 1 0 ^ {- 9}) ^ {2}} = 3. 3 5 \cdot 1 0 ^ {- 9} T.$$

C) The current density of the wire is

$$j = \frac {I}{A} = \frac {I}{\frac {\pi d ^ {2}}{4}}.$$

Thus,

$$d _ {l i m} = 2 \sqrt {\frac {I}{\pi j}} = 2 \sqrt {\frac {0 . 3 0 A}{\pi \cdot 4 0 0 \frac {A}{c m ^ {2}}}} = 0. 0 1 5 c m = 0. 1 5 m m.$$

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