Formulas for motional emf \((E = Blv)\) and the force on a current carrying conductor in a magnetic field \((F = BIl)\). Make sure your units are con sistent
The force on a conductor in a magnetic field is given by:
\( F = I \ell B \)
Where:
The current \( I \) can be expressed as:
\( I = \frac{e}{R} \)
Substitute \( I \) into the force equation:
\[ F = Bv \ell B \cdot \frac{\ell}{R} \]
Simplify:
\[ F = \frac{B^2 \ell^2 v}{R} \]
Given:
Substitute these values into the equation:
\[ F = \frac{(15)^2 \cdot (1)^2 \cdot 4}{5} \]
Simplify:
\[ F = \frac{225 \cdot 4}{5} = 180 \, \text{N} \]
The magnetic force is \( F = 18 \, \text{N}. \)
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: