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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

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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.

This problem has been solved

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