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

Hole Generation in P-type Silicon via Doping

Provided Information:

• Dopant atom to silicon atom ratio: 1 dopant atom per \( 5 \times 10^7 \) silicon atoms.
• Silicon atom density: \( 5 \times 10^{28} \) atoms/m\(^3\).
• Objective: Calculate the number of holes generated per cubic centimetre.

Calculation Process:

Step 1: Dopant to Silicon Atom Ratio

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

Step 2: Dopant Atom Concentration per Cubic Metre

Utilizing the silicon atom density of \( 5 \times 10^{28} \) atoms/m\(^3\), the concentration of dopant atoms per cubic metre is determined: \[ \text{Number of dopant atoms} = \left( \frac{1}{5 \times 10^7} \right) \times (5 \times 10^{28} \, \text{atoms/m}^3) = 10^{21} \, \text{dopant atoms/m}^3 \]

Step 3: Hole Generation Rate per Cubic Metre

Assuming each dopant atom contributes one hole, the number of holes generated per cubic metre directly equals the dopant atom concentration: \[ \text{Number of holes} = 10^{21} \, \text{holes/m}^3 \]

Step 4: Conversion to Holes per Cubic Centimetre

To convert the hole density from cubic metres to cubic centimetres, using the relationship \( 1 \, \text{m}^3 = 10^6 \, \text{cm}^3 \): \[ \text{Number of holes} = \frac{10^{21} \, \text{holes/m}^3}{10^6 \, \text{cm}^3/\text{m}^3} = 10^{15} \, \text{holes/cm}^3 \]

Illustrative Dopant for P-type Silicon:

A common dopant employed to create p-type silicon is boron (B). Boron atoms possess one less valence electron compared to silicon. When boron substitutes a silicon atom in the crystal lattice, it results in the formation of a hole.

Concluding Result:

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