Question

A sinusoidally oscillating current I ( t ) with an amplitude of 4.55 A and a frequency of 879 cycles per second is carried by a long, straight wire. A rectangular loop of copper wire with dimensions b = 45.2 cm by c = 20.3 cm is located a = 50.2 cm from the straight wire, and is coplanar with it. Calculate the average power P avg dissipated by the loop if its resistance is 84.3 Ω .

Answer 1

Answer:

The average power is $4.71\times10^{-8}\ watt$

Explanation:

Given that,

A sinusoidal oscillating current I ( t ) with an amplitude of 4.55 A .

Frequency = 879 cycles

Side b = 45.2 cm

Side c=20.3 cm

Side a = 50.2 cm

Resistance = 84.3 Ω

We need to calculate the angular frequency

Using formula of angular frequency

$\omega=2\pi f$

Put the value into the formula

$\omega=2\times\pi\times879$

$\omega=5522.91\ rad/s$

We need to write the sinusoidal equation

$I(t)=A(\sin\omega t)$

$I(t)=4.55\sin(5522.91 t)$

We need to calculate the magnetic flux

Using formula of flux

$\phi=\int{B\cdot dA}$

$\phi=\int{\dfrac{\mu_{0}I}{2\pi r}dA}$

$\phi=\dfrac{\mu_{0}I}{2\pi}\int_{c}^{a+c}{\dfrac{dr}{r}}$

$\phi=\dfrac{\mu_{0}Ib}{2\pi}ln(\dfrac{a+c}{c})$

$\phi=\dfrac{\mu_{0}b}{2\pi}ln(\dfrac{a+c}{c})(4.55\sin(5522.91 t))$

We need to calculate the induced emf

Using formula of induced emf

$E=-\dfrac{d\phi}{dt}$

$E=-4.55\times\dfrac{\mu_{0}b}{2\pi}ln(\dfrac{a+c}{c})\times5522.91\times\cos(5522.91 t))$

Put the value into the formula

$E=-4.55\times5522.91\times\dfrac{4\pi\times10^{-7}\times0.452}{2\pi}ln(\dfrac{0.502+0.203}{0.203})\times\cos(5522.91 t)$

$E=-0.00282\times\cos(5522.91 t)$

$E=-2.82\times10^{-3}\cos(5522.91 t)$

We need to calculate the average power

Using formula of average power

$P=\dfrac{1}{2}\dfrac{E^2}{R}$

Put the value into the formula

$P=\dfrac{1}{2}\dfrac{(-2.82\times10^{-3})^2}{84.3}$

$P=4.71\times10^{-8}\ watt$

Hence, The average power is $4.71\times10^{-8}\ watt$

Answer 2

The average power dissipated by the given loop if its resistance is 84.3 Ω  is; P_avg = 3.44 × 10⁻⁹ W

We are given;

We are given;

Current; I₀ = 4.55 A

Frequency; f = 879 cycles/s

b = 45.2 cm = 0.452 m

c = 20.3 cm = 0.203 m

a = 50.2 cm = 0.502 m

The formula for the magnetic flux is;

Φ  = $\frac{\mu_{0}*I * c}{2\pi} * In \frac{b + a}{b}}$

Where μ₀ is vacuum permeability with a constant value of 4π × 10⁻⁷ H/m

Meanwhile, the formula for the RMS value of the EMF is;

$E_{rms}$ = $\frac{(\frac{\mu_{0}*c}{2\pi} * In \frac{b + a}{b} * I_{0} * 2\pi f)}{\sqrt{2}}$

Plugging in the relevant values gives;

$E_{rms}$ = $\frac{(\frac{4\pi * 10^{-7}*0.203}{2\pi} * In \frac{0.452 + 0.502}{0.452} * 4.55 * 2\pi *879)}{\sqrt{2}}$

$E_{rms}$ = 5.389 × 10⁻⁴ V

We want to find the average power dissipated by the loop if the resistance is 84.3 Ω .

Formula for power is; P = V²/R

Thus;

P_avg = (5.389 × 10⁻⁴)²/84.3

P_avg = 3.44 × 10⁻⁹ W

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