Answer:
Explanation:
Relation between flux and inductance is as follows
φ = Li
where φ is flux associated with induction of inductance L when a current i flows through it
putting the values
3.25 x 10⁻³ x 800 = L x 2.9
L = .9 H
for induced emf in an induction , the relation is
emf induced = L di / dt
Putting the values
7.5 x 10⁻³ = .9 x di / dt
di / dt = 8.33 x 10⁻³ A / s
(a) The self inductance of the solenoid is 0.897 H.
(b) The magnitude of the rate of change of the current is 0.00836 A/s.
The given parameters;
The self inductance of the solenoid is calculated as follows;
[tex]emf = \frac{d\phi}{dt}\\\\emf = \frac{Ldi}{dt} \\\\d\phi = Ldi\\\\\phi = BA\\\\NBA = LI\\\\L = \frac{NBA}{I} \\\\L = \frac{N\phi}{I} \\\\L = \frac{800 \times 3.25\times 10^{-3}}{2.9} \\\\L = 0.897 \ H\\\\[/tex]
The magnitude of the rate of change of the current is calculated as follows;
[tex]emf = L \frac{di}{dt} \\\\\frac{di}{dt} \ = \frac{emf}{L} \\\\\frac{di}{dt} = \frac{7.5 \times 10^{-3}}{0.897} \\\\\frac{di}{dt} = 0.00836 \ A/s[/tex]
Learn more here:https://brainly.com/question/17086348
