A loop of area 10 cm2 is placed in a uniform magnetic field of 0.5 T with its plane perpendicular to the field. the resistance of the loop is 5 Ω. The loop is suddenly removed from the field. The total charge flown through the loop is 10–x C. Find x.
Question
A loop of area 10 cm2 is placed in a uniform magnetic field of 0.5 T with its plane perpendicular to the field. the resistance of the loop is 5 Ω. The loop is suddenly removed from the field. The total charge flown through the loop is 10–x C. Find x.
Solution
The problem involves Faraday's law of electromagnetic induction, which states that the induced electromotive force (EMF) in any closed circuit is equal to the rate of change of the magnetic flux through the circuit.
Step 1: Calculate the initial magnetic flux. The magnetic flux (Φ) is given by the product of the magnetic field strength (B), the area (A) through which it passes, and the cosine of the angle (θ) between the field lines and the normal (perpendicular) to the area. In this case, the field is perpendicular to the area, so θ = 0 and cos(θ) = 1. Therefore, Φ = B * A = 0.5 T * 10 cm^2 = 0.5 T * 10^-2 m^2 = 5 * 10^-3 Wb (Weber).
Step 2: Calculate the change in magnetic flux. The loop is removed from the field, so the final magnetic flux is 0. Therefore, the change in magnetic flux (ΔΦ) is the final flux minus the initial flux, which is -5 * 10^-3 Wb.
Step 3: Calculate the induced EMF. According to Faraday's law, the induced EMF (ε) is equal to the negative rate of change of the magnetic flux. Since the change happens "suddenly", we can consider it as happening in a very short time, so the rate of change of flux is simply the change in flux. Therefore, ε = -ΔΦ = -(-5 * 10^-3 Wb) = 5 * 10^-3 V.
Step 4: Calculate the total charge flown. The EMF drives a current through the loop, and the amount of charge (Q) that flows is given by Q = I * t, where I is the current and t is the time. Ohm's law tells us that I = ε / R, where R is the resistance. Therefore, Q = ε * t / R. We don't know the time, but we know that the total charge flown is 10^-x C. Therefore, we can write 10^-x C = ε * t / R.
Step 5: Solve for x. We can rearrange the equation to solve for x: x = -log10(ε * t / R). We don't know t, but we know that ε = 5 * 10^-3 V and R = 5 Ω, so we can substitute these values in: x = -log10(5 * 10^-3 V * t / 5 Ω) = -log10(t * 10^-3 C). Therefore, x is the negative logarithm (base 10) of the time in milliseconds that it takes for the loop to be removed from the field.
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