A flip coil is a device used to measure a magnetic field. A coil of radius $r, N\( turns, and electrical resistance \)R$ is initially perpendicular to a magnetic field of magnitude B. The coil is connected to a special kind of galvanometer that measures the total charge \(Q\) that flows through it. To measure the field, the flip coil is rapidly flipped upside down. (a) What is the change in magnetic flux through the coil in one flip? (b) If the time interval during which the coil is flipped is \(\Delta t,\) what is the average induced emf in the coil? (c) What is the average current that flows through the galvanometer? (d) What is the total charge \(Q\) in terms of \(r, N, R,\) and \(B ?\)

Step by step solution

01

(a) Change in magnetic flux during one flip

We start by calculating the initial and final magnetic flux. The magnetic flux through the coil is given by the formula \(\Phi = NBA\cos\theta\), where \(N\) is the number of turns, \(B\) is the magnetic field, \(A\) is the area of the coil, and \(\theta\) is the angle between the magnetic field and the coil's normal vector. Initially, when the coil is perpendicular to the magnetic field, we have \(\theta_i = 0°\). And finally, when it is upside down, \(\theta_f = 180°\). So we calculate the initial and final magnetic flux: \(\Phi_i = NBA\cos(0°) = NBA\) \(\Phi_f = NBA\cos(180°) = -NBA\) Now, we can find the change in magnetic flux during one flip: \(\Delta \Phi = \Phi_f - \Phi_i = -NBA - NBA = -2NBA\)

02

(b) Average induced EMF

According to Faraday's law, the induced EMF in the coil is given by the absolute value of the rate of change in magnetic flux. We'll find the average induced EMF by dividing the change in magnetic flux by the time interval it takes to flip the coil: \(|\varepsilon| = \frac{|\Delta \Phi |}{\Delta t} = \frac{2NBA}{\Delta t}\)

03

(c) Average current

Now, we can determine the average current in the coil by applying Ohm's law, which states that the current is given by the quotient between the induced EMF and the coil's resistance: \(I_{avg}=\frac{|\varepsilon|}{R} = \frac{2NBA}{R\Delta t}\)

04

(d) Total charge Q

Finally, we can find the total charge Q that flows through the coil by multiplying the average current by the time interval: \(Q = I_{avg} \cdot \Delta t = \frac{2NBA}{R\Delta t} \cdot \Delta t\) We notice that the factor \(\Delta t\) will cancel out, leaving us with \(Q = \frac{2NBA}{R}\) Therefore, the total charge Q is given by \(\frac{2NBA}{R}\), where N is the number of turns, B is the magnetic field, A is the area of the coil, and R is its resistance.

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