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 $10.0MHz$. At this field strength, the Hall voltage on the probe is$0.543mV$when the current through the probe is$0.150mA$. Later, when an unknown magnetic field is measured, the Hall voltage at the same current is$1.735mV$. What is the strength of this magnetic field?

Short answer

The strength of the magnetic field is determined as$2.1T$

Step-by-step solution

 Step :1 Introduction

The area of space near a magnetic body or a current-carrying body where magnetic forces caused by the body or current can be detected. The frequency of a charged particle traveling perpendicular to the direction of a uniform magnetic field B is known as the cyclotron frequency or gyrofrequency (constant magnitude and direction). The given is the adjustment of a magnetic field until the proton's cyclotron frequency is $10.0mHz$

The objective is to find the strength of the magnetic field

Step :2 Cyclotron motion 

The cyclotron motion is defined as a particle traveling in a uniform circular motion perpendicular to the magnetic field at a constant speed. The magnetic field causes the circular motion to produce a circular frequency, which is given by equation $(29.21)$in the form

$f_{cyc}=\frac{qB}{2\mathrm{\pi m}}$

Where $q$is the particle's charge, $B$is the magnetic field, and m is the particle's mass. The frequency is determined by the magnetic field rather than the particle's velocity, as indicated by equation $\left(1\right)$. We rewrite equation $\left(1\right)$so that $B$has the form

$B=\frac{2{\mathrm{\pi mf}}_{\mathrm{cyc}}}{q}$

At a frequency of $10MHz$, the Hall voltage $∆v_{1H}=0.543mV$. As a result, we may use this frequency to determine the initial maegtic field B1. Fill in the values for $f,mcyc,andq$from the equation.

$$\begin{gathered} B=\frac{2{\mathrm{\pi mf}}_{\mathrm{cyc}}}{q} \\ =\frac{2\mathrm{\pi }(1.67\times {10}^{-27}\mathrm{kg})(10\times {10}^6\mathrm{Hz})}{1.6\times {10}^{-16}C} \\ =0.655T \end{gathered}$$

Step :3 Potential difference 

The steady-state potential difference between the two surfaces of a conductor is known as the Hall voltage. Equation $(29.25)$in the form relates the Hall voltage to the current $I$inside the conductor and the magnetic field.

$∆V_H=\frac{IB}{tne}$

Where n is the charge density, $t$is the conductor thickness, $e$is the electron charge, and $B$is the applied magnetic field. The Hall voltage is exactly proportional to the magnetic field, as stated by equation $\left(1\right)$.

$∆V_H\alpha B$

We can get an equation for two instants between the Hall voltage and the magnetic field in the form when the current for both probes is the same.

$$\begin{gathered} \frac{∆V_{1\times H}}{∆V_{2\times H}}=\frac{B_1}{B_2} \\ {B}_2=\left(\frac{∆V_{2\times H}}{∆V_{1\times H}}\right)B_1 \end{gathered}$$

We can calculate $B_2$at $∆V_{2H}=1.735mV$by multiplying 1H=0.543 mV by $B1=0.655T$.

$$\begin{gathered} B_2=\left(\frac{∆V_{2\times H}}{∆V_{1\times H}}\right)B_1 \\ =\left(\frac{1.735mV}{0.543mV}\right)(0.655T) \\ =2.1T \end{gathered}$$