Question

A Hall-effect probe to measure magnetic field strengths needs to be calibrated in a known magnetic field. Although it is not easy to do, magnetic fields can be precisely measured by measuring the cyclotron frequency of protons. A testing laboratory adjusts a magnetic field until the proton's cyclotron frequency is 9.60 MHz . At this field strength, the Hall voltage on the probe is 0.540 mV when the current through the probe is 0.143 mA . Later, when an unknown magnetic field is measured, the Hall voltage at the same current is 1.736 mV .A) What is the strength of this magnetic field?

Answer 1

Answer:

2.02395 T

Explanation:

q = Charge = $1.6\times 10^{-19}\ C$

m = Rest mass of proton = $1.67\times 10^{-27}\ kg$

f = Frequency = 9.6 MHz

I = Current

Cylclotron frequency is given by

$f=\frac{qB}{2\pi m}$

Magnetic field is given by

$B=\frac{2\pi mf}{q}\\\Rightarrow B=\frac{2\pi 1.67\times 10^{-27}\times 9.6\times 10^6}{1.6\times 10^{-19}}\\\Rightarrow B=0.62957\ T$

Hall voltage is given by

$v_h=\frac{IBD}{neA}$

For the two cases

$v_h_1=\frac{IB_1D}{neA}$

$v_h_2=\frac{IB_2D}{neA}$

Divinding the Hall voltages

$\frac{v_h_1}{v_h_2}=\frac{B_1}{B_2}\\\Rightarrow B_2=\frac{v_h_2}{v_h_1}\times B_1\\\Rightarrow B_2=\frac{1.736}{0.54}\times 0.62957\\\Rightarrow B_2=2.02395\ T$

The strength of the magnetic field is 2.02395 T

Answer 2

The magnetic field strength of the unknown magnetic field in the later case is 2.02 T.

The given parameters;

• proton's cyclotron frequency, f =  9.60 MHz
• Hall voltage, V = 0.54 mV
• current, I = 0.143 mA
• second hall voltage, V₂ = 1.736 mV

The initial magnetic field strength of the protons is calculated as follows;

$f = \frac{qB}{2\pi m_p} \\\\B = \frac{2\pi m_p f}{q} \\\\B_1 = \frac{2\pi \times1.67 \times 10^{-27}\times 9.6 \times 10^{6} }{1.602 \times 10^{-19}} \\\\B_1 = 0.629 \ T$

The hall voltage is calculated as follows;

$V_H= \frac{IB}{qnd} \\\\V_H_1 = \frac{I B_1}{qnd} \\\\V_H_2 = \frac{I B_2}{qnd} \\\\\frac{V_H_1}{V_H_2} = \frac{B_1}{B_2} \\\\\frac{B_2}{B_1} = \frac{V_H_2}{V_H_1} \\\\B_2 = B_1 (\frac{V_H_2}{V_H_1} )\\\\B_2 = 0.629 \times (\frac{1.736}{0.54} )\\\\B_2 = 2.02 \ T$

Thus, the magnetic field strength in the later case is 2.02 T.

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