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fomenos
2 years ago
14

Inductive charging is used to wirelessly charge electronic devices ranging from toothbrushes to cell phones. Suppose the base un

it of an inductive charger produces a 1.50 ✕ 10−3 T magnetic field. Varying this magnetic field magnitude changes the flux through a 16.0-turn circular loop in the device, creating an emf that charges its battery. Suppose the loop area is 2.75 ✕ 10−4 m2 and the induced emf has an average magnitude of 5.50 V. Calculate the time required (in s) for the magnetic field to decrease to zero from its maximum value.
Physics
1 answer:
denpristay [2]2 years ago
8 0

Answer:

1.2*10^{-6}s

Explanation:

The problem must be addressed through the concepts of electromotive force. By Faraday's law it is defined that

\epsilon = NA \frac{dB}{dt}

Where

\epsilon = Electromotive Force

N = Number of Loops

A = Area

B = Magnetic Field (chaging through the time)

From this equation and our values, we need to find the time, then we re-arrange the equation

dt = NA \frac{dB}{\epsilon}

t = (16)(2.75*10^{-4})\frac{1.50*10^{-3}}{5.50}

t = 1.2*10^{-6}s

Therefore the time required for the magnetic field to decrease to zero from its maximum value is 1.2*10^{-6}s

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The expressions for e/m and the relative error of e/m due to all of the parameters measured:
bija089 [108]

Answer:

Term 1 = (0.616 × 10⁻⁵)

Term 2 = (7.24 × 10⁻⁵)

Term 3 = (174 × 10⁻⁵)

Term 4 = (317 × 10⁻⁵)

(σ ₑ/ₘ) / (e/m) = (499 × 10⁻⁵) to the appropriate significant figures.

Explanation:

(σ ₑ/ₘ) / (e/m) = (σᵥ /V)² + (2 σᵢ/ɪ)² + (2 σʀ /R)² + (2 σᵣ /r)²

mean measurements

Voltage, V = (403 ± 1) V,

σᵥ = 1 V, V = 403 V

Current, I = (2.35 ± 0.01) A

σᵢ = 0.01 A, I = 2.35 A

Coils radius, R = (14.4 ± 0.3) cm

σʀ = 0.3 cm, R = 14.4 cm

Curvature of the electron trajectory, r = (7.1 ± 0.2) cm.

σᵣ = 0.2 cm, r = 7.1 cm

Term 1 = (σᵥ /V)² = (1/403)² = 0.0000061573 = (0.616 × 10⁻⁵)

Term 2 = (2 σᵢ/ɪ)² = (2×0.01/2.35)² = 0.000072431 = (7.24 × 10⁻⁵)

Term 3 = (2 σʀ /R)² = (2×0.3/14.4)² = 0.0017361111 = (174 × 10⁻⁵)

Term 4 = (2 σᵣ /r)² = (2×0.2/7.1)² = 0.0031739734 = (317 × 10⁻⁵)

The relative value of the e/m ratio is a sum of all the calculated terms.

(σ ₑ/ₘ) / (e/m)

= (0.616 + 7.24 + 174 + 317) × 10⁻⁵

= (498.856 × 10⁻⁵)

= (499 × 10⁻⁵) to the appropriate significant figures.

Hope this Helps!!!

6 0
2 years ago
A child blows a leaf straight up in the air from rest. The leaf accelerates at 1.0\,\dfrac{\text m}{\text s^2}1.0 s 2 m1, point,
Neko [114]

Answer:

Time taken by the leaf to displace by 1.0 m distance is

t = \sqrt2 seconds

Explanation:

As we know that initial velocity of the leaf is given as

v_i = 0

now the acceleration upwards for the leaf is

a = 1 m/s^2

The displacement of leaf in upward direction is

d = 1 m

so now we have

d = v_i t + \frac{1}{2}at^2

1 = 0 + \frac{1}{2}(1) t^2

t = \sqrt2 seconds

3 0
2 years ago
Read 2 more answers
Two astronauts of identical mass are connected by a taut cable of negligible mass, as shown in the figure above, and are initial
PolarNik [594]

Answer:

The right answer is "The center of mass doesn't move".

Explanation:

  • It generates a voltage throughout the cable while the astronaut falls on either the wire. At other ends of the spectrum or cable, the tension will be similar. As such, with both astronauts, there would be the same energy, although throughout the opposite way.
  • Thus, the net force seems to be essentially negative on the machine. And therefore the mass center stays stationary.
5 0
2 years ago
Suppose the initial position of an object is zero, the starting velocity is 3 m/s and the final velocity was 10 m/s. The object
Tatiana [17]

Answer:

C. the area of the rectangle plus the area of the triangle under the line

Explanation:

Based on the information provided, the velocity vs. time graph is a line with a positive slope and a y-intercept of (0, 3).  The displacement is the area under this line.  This area can be divided into a triangle and a rectangle.  So of the options available, C is the correct one.

6 0
2 years ago
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If a metallic wire of cross sectional area 3.0 ´ 10-6 m2 carries a current of 6.0 A and has a mobile charge density of 4.24 ´ 10
elena-14-01-66 [18.8K]

Answer:

The drift velocity is v_d=2.9\times 10^{-4}\ m/s.

Explanation:

Given :

Area of metallic wire, A = 3\times 10^{-6}\ m^2.

Current through wire , I=6 \ A.

Mobile charge density , n=4.24\times 10^{28} \ carriers/m^3.

Charge value , e=1.6\times 10^{-19}\ C.

We need to find drift velocity , v_d.

Now, we know :

I=neAv_d

Therefore, v_d=\dfrac{I}{neA}

Putting all given values in above equation we get,

v_d=\dfrac{6}{4.24\times 10^{28}\times 1.6\times 10^{-19} \times 3 \times 10^{-6}}

v_d=2.9\times 10^{-4}\ m/s.

Hence, this is the required solution.

8 0
2 years ago
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