If the current in the toroidal solenoid increases uniformly from zero to 6.0 A in 3.0 μs. Self-inductance of the toroidal solenoid is 40 μH. The magnitude of self-induced emf is
Question
If the current in the toroidal solenoid increases uniformly from zero to 6.0 A in 3.0 μs. Self-inductance of the toroidal solenoid is 40 μH. The magnitude of self-induced emf is
Solution
The self-induced emf in a solenoid can be calculated using Faraday's law of electromagnetic induction, which states that the induced emf is equal to the rate of change of magnetic flux. In the case of a solenoid, this can be expressed as:
emf = -L * (ΔI/Δt)
where:
- L is the self-inductance of the solenoid,
- ΔI is the change in current, and
- Δt is the change in time.
Given that:
- L = 40 μH = 40 * 10^-6 H,
- ΔI = 6.0 A - 0 A = 6.0 A, and
- Δt = 3.0 μs = 3.0 * 10^-6 s,
we can substitute these values into the equation to find:
emf = -40 * 10^-6 H * (6.0 A / 3.0 * 10^-6 s) = -80 V.
However, since the question asks for the magnitude of the self-induced emf, we take the absolute value to get:
emf = 80 V.
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