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Q15P

Expert-verifiedFound in: Page 896

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A square wire loop with 2.00m** **sides is perpendicular to a uniform magnetic field, with half the area of the loop in the field as shown in Figure. The loop contains an ideal battery with emf**** ${\mathit{\epsilon}}{\mathbf{=}}{\mathbf{20}}{\mathbf{.}}{\mathbf{0}}{\mathit{V}}$. If the magnitude of the field varies with time according to**** ${\mathbf{}}{\mathbf{B}}{\mathbf{=}}{\mathbf{0}}{\mathbf{.}}{\mathbf{0420}}{\mathbf{-}}{\mathbf{0}}{\mathbf{.}}{\mathbf{870}}{\mathbf{t}}$, with B in Tesla and t in seconds, (a)what is the net emf in the circuit?(b)what is the direction of the (net) current around the loop?**

a) The net emf in the circuit is,$\epsilon =21.74\mathrm{V}$

b) Direction of the net current around the loop is counterclockwise.

i) Sides of the square loop,$L=2.00\mathrm{m}$

ii) The emf of an ideal battery, $\epsilon =20.0V$

iii) Magnetic field $B=0.0420-0.870t$

**In a square wire loop of 2 m sides and having emf 20 V, the half area is under magnetic field, therefore, the amount of emf in a square covered under magnetic field can be calculated using Faraday’s law. Also from that, the net emf in the circuit can be found. **

**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 follow:

$\epsilon =\frac{-d{\phi}_{B}}{dt}\phantom{\rule{0ex}{0ex}}{\phi}_{B}=BA$

Where,** $\varphi $** is magnetic flux, *B* is magnetic field, *A* is area, *𝜀* is emf.

To calculate net emf in the circuit, first calculate magnetic flux $\phi $.

Let L be the length of a side of the square circuit.

So the area of the square loop will be ${L}^{2}$ Since, only half of the area of the loop is in the field, consider only half of the area to calculate the magnetic flux.

Thus, $A={L}^{2}/2$

As the square loop is perpendicular to the uniform magnetic field, the magnetic flux through the circuit is,

${\phi}_{B}=BA={L}^{2}B/2$

Now, find the induced emf by using Faraday’s law.

${\epsilon}_{i}=\frac{-d{\phi}_{B}}{dt}\phantom{\rule{0ex}{0ex}}{\epsilon}_{i}=-\frac{{L}^{2}}{2}\frac{dB}{dt}$

Now, magnetic field is given as,$B=0.0420-0.870t$

On differentiating,

$\frac{dB}{dt}=-0.870T/s$

Substituting the values in the equation for induced emf,

${\epsilon}_{i}=-\frac{(2.00{)}^{2}}{2}(-0.870)\phantom{\rule{0ex}{0ex}}{\epsilon}_{i}=1.74V$

As the magnetic field is out of the page and decreasing, so the induced emf is counterclockwise around the circuit in the same direction as the emf of the battery.

The total (net) emf is the sum of the emf from the battery and the induced emf.

Therefore,

$\epsilon +{\epsilon}_{i}=20V+1.74V\phantom{\rule{0ex}{0ex}}\epsilon +{\epsilon}_{i}=21.74V$

Hence, the net emf in the circuit is 21.74 V

The net current is in the sense of the total emf i.e. in counterclockwise direction.

Hence, direction of the net current around the loop is counterclockwise.

Therefore, the net emf in the square wire loop, which is partially covered with the uniform magnetic field, can be found by using Faraday’s law.

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