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Q13P

Expert-verifiedFound in: Page 896

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 ${\mathbf{1}}{\mathbf{.}}{\mathbf{20}}{\mathbf{\times}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{3}}{{\mathbf{m}}}^{{\mathbf{2}}}$****. 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, $\mathrm{q}\left(\mathrm{t}\right)=2.95\times {10}^{-2}\mathrm{C}$

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

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

** **

**Faraday's*** law of electromagnetic induction *

Formulae are as follows:

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

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

According to Faraday’s law,

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

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 $\overrightarrow{B}$ is perpendicular to the area A, then,

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

Therefore,

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

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

** **

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.

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