Q. 5: (a)The Hall effect can be used to calculate the charge-carrier number density in a di0 2i 1ed conductor. If a conductor carrying a current of 2.0 A is 0.5 mm thick, and the Hall effect bognvoltage is 4.5µV when it is in a uniform magnetic field of 1.2 T, what is the density of lotar charge carriers in the conductor? bo sdiaAboen on
Q. 5: (a)The Hall effect can be used to calculate the charge-carrier number density in a di0 2i 1ed conductor. If a conductor carrying a current of 2.0 A is 0.5 mm thick, and the Hall effect bognvoltage is 4.5µV when it is in a uniform magnetic field of 1.2 T, what is the density of lotar charge carriers in the conductor? bo sdiaAboen on
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![**Question 5: (a)** The Hall effect can be used to calculate the charge-carrier number density in a conductor. If a conductor carrying a current of 2.0 A is 0.5 mm thick, and the Hall effect voltage is 4.5 μV when it is in a uniform magnetic field of 1.2 T, what is the density of charge carriers in the conductor?
*Explanation:*
The Hall effect is a phenomenon in which a voltage difference (Hall voltage) is produced across an electrical conductor when it is placed in a magnetic field perpendicular to the current. This effect can be used to determine the density of charge carriers (such as electrons) in the conductor.
To calculate the charge-carrier number density (\( n \)), the Hall effect equation can be utilized:
\[ V_H = \frac{IB}{nq} \]
Where:
- \( V_H \) is the Hall voltage (4.5 μV),
- \( I \) is the current (2.0 A),
- \( B \) is the magnetic field (1.2 T),
- \( q \) is the charge of an electron (\( 1.6 \times 10^{-19} \) C).
The thickness of the conductor is used in calculations to determine the area through which current flows. The number density \( n \) can be rearranged and calculated as follows:
\[ n = \frac{IB}{V_H q} \]
**Graph/Diagram Explanation:**
There is a faint diagram that resembles a simple circuit with a resistor and other components. However, due to its poor visibility, specific details about the diagram cannot be accurately described.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe4d256a5-cbd6-4aee-ae86-020aabe91708%2F3234ca72-fd09-4b56-aacf-f885374f0985%2F67van2f_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 5: (a)** The Hall effect can be used to calculate the charge-carrier number density in a conductor. If a conductor carrying a current of 2.0 A is 0.5 mm thick, and the Hall effect voltage is 4.5 μV when it is in a uniform magnetic field of 1.2 T, what is the density of charge carriers in the conductor?
*Explanation:*
The Hall effect is a phenomenon in which a voltage difference (Hall voltage) is produced across an electrical conductor when it is placed in a magnetic field perpendicular to the current. This effect can be used to determine the density of charge carriers (such as electrons) in the conductor.
To calculate the charge-carrier number density (\( n \)), the Hall effect equation can be utilized:
\[ V_H = \frac{IB}{nq} \]
Where:
- \( V_H \) is the Hall voltage (4.5 μV),
- \( I \) is the current (2.0 A),
- \( B \) is the magnetic field (1.2 T),
- \( q \) is the charge of an electron (\( 1.6 \times 10^{-19} \) C).
The thickness of the conductor is used in calculations to determine the area through which current flows. The number density \( n \) can be rearranged and calculated as follows:
\[ n = \frac{IB}{V_H q} \]
**Graph/Diagram Explanation:**
There is a faint diagram that resembles a simple circuit with a resistor and other components. However, due to its poor visibility, specific details about the diagram cannot be accurately described.
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