Understanding the Problem
We are given the induced emf in a coil and the rate of change of current, and we need to find the inductance of the coil.
Solution
1. Formula for Induced EMF:
The induced emf in a coil is given by:
\( e = -L \frac{di}{dt} \)
where:
2. Given Values:
3. Substitute Values into the Formula:
\( 20 = -L \times (-2 \times 10^3) \)
4. Solve for Inductance (L):
\( L = \frac{20}{2 \times 10^3} \)
\( L = 10 \times 10^{-3} \, \text{H} \)
\( L = 10 \, \text{mH} \)
Final Answer
The inductance of the coil is \( 10 \, \text{mH} \).
A wire of resistance $ R $ is bent into a triangular pyramid as shown in the figure, with each segment having the same length. The resistance between points $ A $ and $ B $ is $ \frac{R}{n} $. The value of $ n $ is:


Let \( a \in \mathbb{R} \) and \( A \) be a matrix of order \( 3 \times 3 \) such that \( \det(A) = -4 \) and \[ A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix} \] where \( I \) is the identity matrix of order \( 3 \times 3 \).
If \( \det\left( (a + 1) \cdot \text{adj}\left( (a - 1) A \right) \right) \) is \( 2^m 3^n \), \( m, n \in \{ 0, 1, 2, \dots, 20 \} \), then \( m + n \) is equal to: