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Found in: Page 896

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# One hundred turns of (insulated) copper wire are wrapped around a wooden cylindrical core of cross-sectional area ${\mathbf{1}}{\mathbf{.}}{\mathbf{20}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{3}}{{\mathbf{m}}}^{{\mathbf{2}}}$. The two ends of the wire are connected to a resistor. The total resistance in the circuit is ${\mathbf{13}}{\mathbf{.}}{\mathbf{0}}{\mathbf{\Omega }}$. If an externally applied uniform longitudinal magnetic field in the core changes from 1.60 T in one direction to 1.60 T in the opposite direction, how much charge flows through a point in the circuit during the change?

The charge flowing through a point in the circuit during the change in magnetic field is, $\mathrm{q}\left(\mathrm{t}\right)=2.95×{10}^{-2}\mathrm{C}$

See the step by step solution

## Step 1: Given

1. Cross-sectional area, $\mathrm{A}=1.20×{10}^{-3}{\mathrm{m}}^{2}$
2. Resistance, $R=13.0\Omega$
3. Magnetic field at core, B (0) = 1.60 T
4. Magnetic field at other end in opposite direction, B (t) = (-1.60 T)

## Step 2: Determining the concept

The magnetic flux ${{\mathbit{\phi }}}_{{\mathbit{B}}}$ through area A in a magnetic field is given by Faraday’s law of induction. If the magnetic field $\stackrel{\mathbf{\to }}{\mathbf{B}}$ is perpendicular to the area A, then substitute the given values in the formula to find the charge.

Faraday's law of electromagnetic induction states, Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced in it.

Formulae are as follows:

$emf=-N.\frac{d\phi }{dt}$

Where, $d\Phi$ is magnetic flux, N is number of turns, dt is time.

## Step 3: Determining the charge flowing through a point in the circuit during the change in magnetic field

$emf=-N.\frac{d\phi }{dt}$

From Ohm’s law, emf = i.R and i = dq/dt

$\therefore i.R=-N.\frac{d\phi }{dt}$

$\frac{dq}{dt}=-N.\frac{d\phi }{dt}$

Integrating this with respect to time,

$q{\left(t\right)}_{0}^{t}=-N\phi {\left(t\right)}_{0}^{t}$

To calculate the charge flowing through the point,

$q\left(t\right)=\frac{N}{R}\left[{\phi }_{B}\left(0\right)-{\phi }_{B}\left(t\right)\right]$

According to Faraday’s law, if the magnetic field $\stackrel{\to }{B}$ is perpendicular to the area A, then,

${\phi }_{B}=BA$

Therefore,

$\mathrm{q}\left(\mathrm{t}\right)=\frac{\mathrm{NA}}{\mathrm{R}}\left[\mathrm{B}\left(0\right)-\mathrm{B}\left(\mathrm{t}\right)\right]\phantom{\rule{0ex}{0ex}}\mathrm{q}\left(\mathrm{t}\right)=\frac{100×\left(1.20×{10}^{-3}{\mathrm{m}}^{2}\right)}{\left(13.0\mathrm{\Omega }\right)}\left[16.0\mathrm{T}-\left(-1.60\mathrm{T}\right)\right]\phantom{\rule{0ex}{0ex}}\mathrm{q}\left(\mathrm{t}\right)=100×\left(0.0923×{10}^{-3}{\mathrm{m}}^{2}\right)×\left(3.20\right)\phantom{\rule{0ex}{0ex}}\mathrm{q}\left(\mathrm{t}\right)=2.95×{10}^{-2}\mathrm{C}$

Hence, the charge flowing through a point in the circuit during the change in magnetic field is, $q\left(t\right)=2.95×{10}^{-2}\mathrm{C}$

Therefore, the charge flow through a point in the circuit during the change in the magnetic field can be found by using Faraday’s law.