Irodov 3.299: Charge Counts the Flux Change – Amit & Deepali Sinha Foundation

Problem Irodov 3.299

A small coil of $N = 60$ turns, each of cross-sectional area $S = 3.0$ mm $^2$ , sits in the uniform field between the poles of an electromagnet with its axis along the field. The coil is connected to a ballistic galvanometer, and the whole circuit has resistance $R = 40$ $\Omega$ . The coil is turned through $180^\circ$ about one of its diameters, and the galvanometer registers a charge $q = 4.5$ $\mu$ C passing through it. Find the magnitude of the magnetic induction between the poles.

Before: the coil's normal n along the uniform field B. — Before: the coil’s normal $\vect n$ along the uniform field $\vect B$ .

After the half turn about a diameter: n points against B. — After the half turn about a diameter: $\vect n$ points against $\vect B$ .

Given

$S = 3.0$ mm $^2$
$N = 60$
turn through $180^\circ$
$q = 4.5$ $\mu$ C
$R = 40$ $\Omega$

Find

B

Solution

Step 1The flux linkage before and after

With the normal $\vect n$ fixed to the coil, each turn first carries the flux $BS$ and, after the half turn, $-BS$ , since $\vect n$ then points against the field. The change of the flux linked by all $N$ turns is therefore

\Delta\Psi = N(-BS) - N(BS) = -2NBS \tag{1}

Step 2The charge through the circuit

While the coil turns, Faraday’s law gives an emf $\mathcal{E}_i = -d\Psi/dt$ , and with the total resistance $R$ the current is $i = \mathcal{E}_i/R$ . The charge is the time integral of the current over the whole turn, and with (1):

\begin{aligned} q &= \int i\,dt = -\frac{1}{R}\int\frac{d\Psi}{dt}\,dt = -\frac{\Delta\Psi}{R} \\ &= \frac{2NBS}{R} \end{aligned} \tag{2}

Step 3The field

Solving (2) for $B$ and substituting the givens:

B = \frac{(4.5\times 10^{-6}\text{ C})(40\ \Omega)}{2\cdot 60\cdot(3.0\times 10^{-6}\text{ m}^2)}

B = \frac{qR}{2NS} = 0.50\text{ T}

(3)

Check

Units: $\text{C}\cdot\Omega = \text{V}\cdot\text{s} = \text{Wb}$ , and dividing by an area in $\text{m}^2$ leaves $\text{Wb/m}^2 = \text{T}$ . Size: the flux through one turn is $BS = 1.5\times 10^{-6}$ Wb, the linked flux $N\cdot BS = 9.0\times 10^{-5}$ Wb reverses, a change of $1.8\times 10^{-4}$ Wb, and dividing by $40\ \Omega$ gives back $4.5\ \mu$ C.

Remark

A quarter turn, from along the field to across it, would pass only half this charge: the flux goes from $NBS$ to zero. The half turn doubles the reading by reversing the flux, which is why search coils are flipped rather than merely pulled out of the field (pulling the coil out also gives $NBS$ , just like the quarter turn).