Chapter 20, Problem 19P

To determine

(a) To determine:

The unknown value of magnetic field in a Hall probe when $\theta ={90}^o$

Answer to Problem 19P

Solution:

The value of magnetic field in a Hall probe is 0.525T if $\theta ={90}^o$

Explanation of Solution

Hall emf is proportional to magnetic field, by finding the ratio of the two given cases, the unknown magnetic field can be calculated.

Given:

$\begin{array}{l}\text{When}\theta =\text{9}{0}^{\text{o}},B=0.\text{1T},V_{hall}=\text{ 12mV}. \\ \left(\text{a}\right)\text{ When}\theta =\text{9}{0}^{\text{o}},B=? \text{If }{V}_{hall}=\text{ 63mV }\left(\text{b}\right)\text{ When}\theta =\text{6}{0}^{\text{o}},B=? \text{If }{V}_{hall}=\text{ 63mV}.\end{array}$

Formula used:

The magnitude of Hall emf is given by,

$V_{Hall}=d{v}_{d}B\sin \theta$,

where, d is the width of the conductor,

B is the strength of the magnetic field,

v_d is the drift velocity of electron,

ϴ is the angle between magnetic field and normal to the conductor

Calculation:

$V_{Hall}=d{v}_{d}B\sin \theta$

$V_{Hall}\propto B\sin \theta$

$\frac{V_2}{V_1}=\frac{B_2\sin {\theta }_2}{B_1\sin {\theta }_1}$

When $\theta =\text{9}{0}^{\text{o}},B=? \text{If }{V}_{hall}=\text{ 63mV}$

$\frac{63}{12}=\frac{B_2\sin 90}{0.1\times \sin 90}$

$B_2=0.525T$

To determine

(b) To determine:

The unknown value of magnetic field in a Hall probe when $\theta ={60}^o$

Answer to Problem 19P

Solution:

The value of magnetic field in a Hall probe is 0.6T if $\theta ={60}^o$

Explanation of Solution

Calculation:

When $\theta ={60}^{\text{o}},B=? \text{If }{V}_{hall}=\text{ 63mV}$

$\frac{63}{12}=\frac{B_2\sin 60}{0.1\times \sin 90}$

$B_2=0.6T$