Question

A current-carrying loop can be considered as a magnetic dipole placed along its axis. Explain.

Hint:

A current-carrying loop generates a magnetic dipole field, and its magnetic dipole moment is given by \( \mathbf{M} = I A \hat{n} \).

Answer 1

Number of Holes Created in P-type Silicon Due to Doping 

Given:

• The average number of dopant atoms per \( 5 \times 10^7 \) silicon atoms
• The number density of silicon atoms in the specimen is \( 5 \times 10^{28} \) atoms/m\(^3\)
• We need to calculate the number of holes created per cubic centimetre due to doping.

Solution:

Step 1: Number of dopant atoms per silicon atom

The ratio of dopant atoms to silicon atoms is given as: \[ \frac{\text{Number of dopant atoms}}{\text{Number of silicon atoms}} = \frac{1}{5 \times 10^7} \]

Step 2: Number of dopant atoms per cubic metre of silicon

The number density of silicon atoms is \( 5 \times 10^{28} \) atoms/m\(^3\). The number of dopant atoms per cubic metre is: \[ \text{Number of dopant atoms} = \frac{1}{5 \times 10^7} \times 5 \times 10^{28} = 10^{21} \, \text{dopant atoms/m}^3 \]

Step 3: Number of holes created per cubic metre

Since each dopant atom creates one hole in the semiconductor, the number of holes created per cubic metre is equal to the number of dopant atoms: \[ \text{Number of holes} = 10^{21} \, \text{holes/m}^3 \]

Step 4: Convert the number of holes to per cubic centimetre

Since \( 1 \, \text{m}^3 = 10^6 \, \text{cm}^3 \), the number of holes per cubic centimetre is: \[ \text{Number of holes} = \frac{10^{21}}{10^6} = 10^{15} \, \text{holes/cm}^3 \]

Example of a Dopant in P-type Silicon:

One example of a dopant used in p-type silicon is **boron (B)**. Boron atoms have one fewer valence electron than silicon, creating a hole in the crystal lattice when substituted for a silicon atom.

Final Answer:

The number of holes created per cubic centimetre in the specimen due to doping is \( \boxed{10^{15}} \, \text{holes/cm}^3 \).