The magnetic flux linked with a coil changes by 2×10-2Wb when the currentchanges by 0.01A. The self inductance of the coil is
Question
The magnetic flux linked with a coil changes by 2×10<sup>-2</sup> Wb when the current changes by 0.01 A. The self inductance of the coil is
Solution
The self-inductance of a coil is defined as the ratio of the magnetic flux to the current. It is given by the formula:
L = ΔΦ/ΔI
where: L is the self-inductance, ΔΦ is the change in magnetic flux, and ΔI is the change in current.
Given that ΔΦ = 2×10^-2 Wb and ΔI = 0.01 A, we can substitute these values into the formula to find the self-inductance:
L = (2×10^-2 Wb) / (0.01 A) = 2 H
So, the self-inductance of the coil is 2 Henry.
Similar Questions
The current in a coil of inductance 5H decreases at the rate of 2A/s. The induced e.m.f is
The current in a coil decreases from 30 amp to 10 amp in 0.2 sec if the self inductance of coil is 4 Henry. What is the induced emf in volts?
A coil induces 350 mV when the current changes at the rate of 1 A/s. The value of inductance is:a.350mHb.250mHc.150mHd.3,500mH
When the current changes from +2 A to - 2 A in 0.05s, and EMF of 8V is inducted in a coil. The coefficient of self-induction of the coil is
Design a 2 cm long “air cored” rod type inductor to have8 µF to within 5%.(b) (1 mark) Calculate the magnetic flux Φ in Wb through the coil from0.75 A DC
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.