Question: At a certain place, Earth’s magnetic field has magnitude$B=0.590\mathrm{gauss}$and is inclined downward at an angle of 70.0$°$to the horizontal. A flat horizontal circular coil of wire with a radius of 10.0 cmhas 1000 turnsand a total resistance of$85.0\Omega$. It is connected in series to a meter with$140\Omega$resistance. The coil is flipped through a half-revolution about a diameter, so that it is again horizontal. How much charge flows through the meter during the flip?

1. Magnitude of the magnetic field, $B=0.590G=0.59\times {10}^{-4}T$
2. Radius of coil wire, R = 10.0cm=0.1 m
3. Number of turns, N = 1000
4. Coil resistance $85.0\Omega$
5. Meter resistance $140\Omega$
6. Earth’s magnetic field is inclined downward at an angle of 70.0$°$ to the horizontal

The current is the rate of change of the charge. This current can be related to Ohm’s law which gives the relation between current and emf to determine the charge flow.

Faraday'slaw of electromagnetic inductionstates, Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced in it.

Ohm's law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperatures remain constant.

Formulae are as follows:

$$\begin{gathered} i=\frac{dq}{dt} \\ i=\frac{\epsilon }{R} \\ \epsilon =-N\frac{d\mathbf{\varphi }}{dt} \\ \mathbf{\varphi }=\int B.dA \end{gathered}$$

Where,is magnetic flux, B is magnetic field, A is area, i is current, R is resistance, 𝜀 is emf, Nis number of turns, q is charge.

The current i and charge q relation is,

$i=\frac{dq}{dt}$

From Ohm’s law,

$i=\frac{\epsilon }{R}$

From Faraday’s law,

$\epsilon =-N\frac{d\mathbf{\varphi }}{dt}$

Hence, by using the above equations,

role="math" localid="1661858329292" $$\begin{gathered} \frac{dq}{dt}=\frac{\epsilon }{R\mathrm{l}} \\ \frac{dq}{dt}=\frac{1}{R}\left(-N\frac{d\mathbf{\varphi }}{dt}\right)=-\frac{N}{R}\left(\frac{d\mathbf{\varphi }}{dt}\right) \\ \frac{dq}{dt}=-\frac{N}{R}\left(\frac{d\mathbf{\varphi }}{dt}\right) \end{gathered}$$

Hence,

$$\begin{gathered} dq=-\frac{N}{R}d\mathbf{\varphi } \\ dq=-\frac{N}{R}d\mathbf{\varphi } \end{gathered}$$

By integrating on both the side,

$$\begin{gathered} q=-\frac{N}{R}\int d\mathbf{\varphi } \\ q=-\frac{N}{R}\int d\left({\mathbf{\varphi }}_{\mathbf{2}}\mathbf{-}{\mathbf{\varphi }}_{\mathbf{1}}\right) \\ q=\frac{N}{R}\int d\left({\mathbf{\varphi }}_{\mathbf{1}}\mathbf{-}{\mathbf{\varphi }}_{\mathbf{2}}\right) \\ q=\frac{N}{R}\left(BA\cos {\theta }_1-BA\cos {\theta }_2\right) \end{gathered}$$

${\theta }_1$is the angle before the coil is flipped, ${\theta }_1=90°-70°=20°$from the axis of earth rotation.

${\theta }_2$is the angle after the coil is flipped,${\theta }_2=20°+180°$from the axis of earth rotation.

$$\begin{gathered} q=\frac{N}{R}\left(BA\cos 20°-BA\cos \left(20°+180°\right)\right) \\ q=\frac{N}{R}\left(BA\cos 20°-\left(-BA\cos 20°\right)\right) \\ q=\frac{2N}{R}\left(BA\cos 20°\right) \\ q=\frac{2\times 1000}{\left(85+140\right)}\left(\left(0.59\times {10}^{-4}\right)\left(\pi \times 0.1^2\right)\cos 20°\right) \\ q=1.55\times {10}^{-5}c \end{gathered}$$

Hence, the charge flows through the meter when the coil is flipped is,$q=1.55\times {10}^{-5}c$.

Therefore, the charge flowing during the coil when it is flipped can be determined as discussed above by considering the earth’s magnetic field.