Question

A solenoid coil.has 10^4 turns of fine insulated conducting wire. The main cross sectional area of the coil is 4000m^2. The magnetic field through the coil changes at a uniform rate of 0.8T to -0.4T in a time of 3s. If the circuit resistance of the coil is 12ohms. Calculate (1).the charge made to pass a cross section of the circuit. (2).the average current (3).the instantaneous current when the magnetic field is zero

Answer 1

Answer on Question #82351, Physics / Electromagnetism

Question:

A solenoid coil has $10^4$ turns of fine insulated conducting wire. The main cross sectional area of the coil is $4000\mathrm{m}^2$. The magnetic field through the coil changes at a uniform rate of 0.8T to -0.4T in a time of 3s. If the circuit resistance of the coil is 12ohms. Calculate (1), the charge made to pass a cross section of the circuit. (2), the average current (3), the instantaneous current when the magnetic field is zero

Solution:

In accordance with Faraday's law $\mathrm{E} = \frac{d\varPhi}{dt}$ and the current $I = \frac{E}{R}$. Then

$$q = \int I dt = \int \frac{d\varPhi}{R dt} dt = \frac{\Delta\varPhi}{R} = \frac{(B_1 - B_2)SN}{R}, \text{ respectively } q = \frac{1.2 \cdot 4000 \cdot 10^4}{12} = 4\,MC.$$

Again the current $I = \frac{E}{R} = \frac{\Delta BSN}{\tau R} = \frac{1.2 \cdot 4000 \cdot 10^4}{36} = 1.33\,MA$, this value is the average and the instantaneous current, simultaneously.

The answer:

The charge $q = \frac{1.2 \cdot 4000 \cdot 10^4}{12} = 4\,MC$

The current $I = \frac{E}{R} = \frac{\Delta BSN}{\tau R} = \frac{1.2 \cdot 4000 \cdot 10^4}{36} = 1.33\,MA$

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