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gulaghasi [49]
3 years ago
10

My Notes A flat loop of wire consisting of a single turn of cross-sectional area 7.80 cm2 is perpendicular to a magnetic field t

hat increases uniformly in magnitude from 0.500 T to 3.20 T in 1.02 s. What is the resulting induced current if the loop has a resistance of 1.30 ?
Physics
1 answer:
Nastasia [14]3 years ago
6 0

Answer:

0.00158 A

Explanation:

A = Area of loop = 7.8 cm²

B_f = Final magnetic field = 3.2 T

B_i = Initial magnetic field = 0.5 T

R = Resitance = 1.3\ \Omega

The emf is given by

E=\frac{(B_f-B_i)A}{t}\\\Rightarrow E=\frac{(3.2-0.5)\times 7.8\times 10^{-4}}{1.02}\\\Rightarrow E=0.00206\ V

The current is given by

I=\frac{E}{R}\\\Rightarrow I=\frac{0.00206}{1.3}\\\Rightarrow I=0.00158\ A

The resulting induced current in the loop is 0.00158 A

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i_2=3.61\ A

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

q, q_1, q_2 = charge of the capacitor in any time t, t_1, t_2

q_o = initial charge of the capacitor

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The charge in an LC circuit is given by

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q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]

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From eq 1:

\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}

From eq 2:

\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}

Squaring and adding the last two equations, and knowing that

sin^2x+cos^2x=1

\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1

Operating

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Solving for q_o

\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}

Now we know the value of q_0, we repeat the procedure of eq 1 and eq 2, but now at the second time t_2, and solve for i_2

\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2

Solving for i_2

\displaystyle i_2=w\sqrt{q_o^2-q_2^2}

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

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w=2\pi f=500\ rad/s

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q_0=0.01166\ c

Finally

\displaystyle i_2=500\sqrt{0.01166^2-.006^2}

i_2=5\ A

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