Chapter 26, Problem 66P

(a)

To determine

The magnetic field

Answer to Problem 66P

  $1.36\text{ T}$

Explanation of Solution

Given:

Number density of electrons $=n=8.47\times {10}^{22}{\text{ cm}}^{-3}=8.47\times {10}^{22}\times {\text{10}}^6{\text{m}}^{-3}=8.47\times {10}^{28}{\text{m}}^{-3}$

Current in the metal strip $=I=20\text{ A}$

Width of the metal strip $=b=2\text{ cm = 0}\text{.02 m}$

Thickness of the metal strip $=t=0.1\text{ cm = 0}\text{.001 m}$

Area of cross-section of the strip $=A$

Magnitude of charge on each electron $=e=1.6\times {10}^{-19}C$

Drift speed of electrons $=v_d$

Hall voltage $=V=2\text{ μV}=2\times {10}^{-6}\text{ V}$

Formula Used:

Area of cross-section of a rectangular shape strip is given as

  $A=bt$

Current is given as

  $I=neA{v}_d$

Hall voltage is given as

  $V=Bb{v}_d$

Calculation:

Area of cross-section of the strip is given as

  $A=bt$

Current flowing through the strip is given as

  $\begin{array}{l}I=neA{v}_d\\ I=ne(bt)v_d\\ v_d=\frac{I}{nebt}\text{ Eq-1}\end{array}$

Hall voltage is given as

  $\begin{array}{l}V=Bb{v}_d\\ \text{using Eq-1}\\ V=Bb\left(\frac{I}{nebt}\right)\\ V=B\left(\frac{I}{net}\right)\\ V(net)=BI\\ B=\frac{neVt}{I}\end{array}$

Inserting the values, we get

  $\begin{array}{l}B=\frac{(8.47\times {10}^{28})(1.6\times {10}^{-19})(2\times {10}^{-6})(0.001)}{20}\\ B=1.36\text{ T}\end{array}$

Conclusion:

The magnetic field comes out to be $1.36\text{ T}$

(b)

To determine

The magnetic field

Answer to Problem 66P

  $3.56\text{ T}$

Explanation of Solution

Given:

Number density of electrons $=n=8.47\times {10}^{22}{\text{ cm}}^{-3}=8.47\times {10}^{22}\times {\text{10}}^6{\text{m}}^{-3}=8.47\times {10}^{28}{\text{m}}^{-3}$

Current in the metal strip $=I=20\text{ A}$

Width of the metal strip $=b=2\text{ cm = 0}\text{.02 m}$

Thickness of the metal strip $=t=0.1\text{ cm = 0}\text{.001 m}$

Area of cross-section of the strip $=A$

Magnitude of charge on each electron $=e=1.6\times {10}^{-19}C$

Drift speed of electrons $=v_d$

Hall voltage $=V=5.25\text{ μV}=5.25\times {10}^{-6}\text{ V}$

Formula Used:

Area of cross-section of a rectangular shape strip is given as

  $A=bt$

Current is given as

  $i=neA{v}_d$

Hall voltage is given as

  $V=Bb{v}_d$

Calculation:

Area of cross-section of the strip is given as

  $A=bt$

Current flowing through the strip is given as

  $\begin{array}{l}I=neA{v}_d\\ I=ne(bt)v_d\\ v_d=\frac{I}{nebt}\text{ Eq-1}\end{array}$

Hall voltage is given as

  $\begin{array}{l}V=Bb{v}_d\\ \text{using Eq-1}\\ V=Bb\left(\frac{I}{nebt}\right)\\ V=B\left(\frac{I}{net}\right)\\ V(net)=BI\\ B=\frac{neVt}{I}\end{array}$

Inserting the values, we get

  $\begin{array}{l}B=\frac{(8.47\times {10}^{28})(1.6\times {10}^{-19})(5.25\times {10}^{-6})(0.001)}{20}\\ B=3.56\text{ T}\end{array}$

Conclusion:

The magnetic field comes out to be $3.56\text{ T}$

(c)

To determine

The magnetic field

Answer to Problem 66P

  $5.42\text{ T}$

Explanation of Solution

Given:

Number density of electrons $=n=8.47\times {10}^{22}{\text{ cm}}^{-3}=8.47\times {10}^{22}\times {\text{10}}^6{\text{m}}^{-3}=8.47\times {10}^{28}{\text{m}}^{-3}$

Current in the metal strip $=I=20\text{ A}$

Width of the metal strip $=b=2\text{ cm = 0}\text{.02 m}$

Thickness of the metal strip $=t=0.1\text{ cm = 0}\text{.001 m}$

Area of cross-section of the strip $=A$

Magnitude of charge on each electron $=e=1.6\times {10}^{-19}C$

Drift speed of electrons $=v_d$

Hall voltage $=V=8\text{ μV}=8\times {10}^{-6}\text{ V}$

Formula Used:

Area of cross-section of a rectangular shape strip is given as

  $A=bt$

Current is given as

  $I=neA{v}_d$

Hall voltage is given as

  $V=Bb{v}_d$

Calculation:

Area of cross-section of the strip is given as

  $A=bt$

Current flowing through the strip is given as

  $\begin{array}{l}I=neA{v}_d\\ I=ne(bt)v_d\\ v_d=\frac{I}{nebt}\text{ Eq-1}\end{array}$

Hall voltage is given as

  $\begin{array}{l}V=Bb{v}_d\\ \text{using Eq-1}\\ V=Bb\left(\frac{I}{nebt}\right)\\ V=B\left(\frac{I}{net}\right)\\ V(net)=BI\\ B=\frac{neVt}{I}\end{array}$

Inserting the values, we get

  $\begin{array}{l}B=\frac{(8.47\times {10}^{28})(1.6\times {10}^{-19})(8\times {10}^{-6})(0.001)}{20}\\ B=5.42\text{ T}\end{array}$

Conclusion:

The magnetic field comes out to be $5.42\text{ T}$