A magnet of length `10 cm` and magnetic moment `1Am^(2)` is placed along the side of an equilateral triangle of the side `AB` of length `10 cm`. The magnetic induction at third vetex `C` is

Options

A. `10^(-9)T`
B. `10^(-7)`
C. `10^(-5)T`
D. `10^(-4)T`

Explanation

To find the magnetic induction at the third vertex \( C \) of an equilateral triangle formed by points \( A \) and \( B \), where a magnet of length \( 10 \, \text{cm} \) and magnetic moment \( 1 \, \text{Am}^2 \) is placed along side \( AB \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Length of the magnet \( L = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Magnetic moment \( M = 1 \, \text{Am}^2 \)
- Side length of the equilateral triangle \( AB = 10 \, \text{cm} = 0.1 \, \text{m} \)

2. **Calculate the Pole Strength:**
- The pole strength \( m \) can be calculated using the formula:
\[ m = \frac{M}{L} \]
- Substituting the values:
\[ m = \frac{1 \, \text{Am}^2}{0.1 \, \text{m}} = 10 \, \text{A} \]

3. **Calculate the Magnetic Field at Point C Due to Pole A:**
- The magnetic field \( B_A \) at point \( C \) due to pole \( A \) is given by:
\[ B_A = \frac{\mu_0}{4\pi} \cdot \frac{m}{r^2} \]
- Here, \( r = 0.1 \, \text{m} \) (distance from pole A to point C).
- The permeability of free space \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).
- Substituting the values:
\[ B_A = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{10}{(0.1)^2} = 10^{-6} \cdot 10^2 = 10^{-4} \, \text{T} \]

4. **Calculate the Magnetic Field at Point C Due to Pole B:**
- The magnetic field \( B_B \) at point \( C \) due to pole \( B \) is also:
\[ B_B = \frac{\mu_0}{4\pi} \cdot \frac{m}{r^2} = 10^{-4} \, \text{T} \]
- Since both poles are equal in strength and distance from point \( C \), \( B_B \) is also \( 10^{-4} \, \text{T} \).

5. **Determine the Resultant Magnetic Field at Point C:**
- The angle between the two magnetic fields \( B_A \) and \( B_B \) is \( 120^\circ \) (since they are at the vertices of an equilateral triangle).
- Using the formula for the resultant of two vectors:
\[ B_{resultant} = \sqrt{B_A^2 + B_B^2 + 2B_A B_B \cos(120^\circ)} \]
- Since \( B_A = B_B = 10^{-4} \, \text{T} \):
\[ B_{resultant} = \sqrt{(10^{-4})^2 + (10^{-4})^2 + 2(10^{-4})(10^{-4})(-0.5)} \]
\[ = \sqrt{2 \times (10^{-4})^2 - (10^{-4})^2} = \sqrt{(10^{-4})^2} = 10^{-4} \, \text{T} \]

6. **Conclusion:**
- The magnetic induction at point \( C \) is:
\[ B = 10^{-4} \, \text{T} \]