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maksim [4K]
3 years ago
13

A 65-cm segment of conducting wire carries a current of 0.35 A. The wire is placed in a uniform magnetic field that has a magnit

ude of 1.24 T. What is the angle between the wire segment and the magnetic field if the force on the wire is 0.26 N

Physics
2 answers:
Artyom0805 [142]3 years ago
8 0

Answer: The angle between the wire segment and the magnetic field 66.42°

Explanation:

Please see the attachment below

Setler [38]3 years ago
8 0

Answer:

67.17°

Explanation:

From the question,

F = BILsinФ............. Equation 1

Where F = force on the wire, B = magnetic field, I = current carried by the wire, L = length of the wire, Ф = angle between the wire segment and the magnetic field.

make Ф the subject of the equation

Ф =sin⁻¹ (F/BIL)............... Equation 2

Given; F = 0.26 N, B = 1.24, I = 0.35 A, L = 65 cm = 0.65 m

Substitute into equation 2

Ф = sin⁻¹[0.26/(1.24×0.35×0.65)]

Ф = sin⁻¹(0.26/0.2821)

Ф = sin⁻¹(0.9217)

Ф = 67.17°

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Explanation:

m = 5 kg

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3 years ago
Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular alumin
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Answer:

\rm 9.186\times 10^{-7}\ C.

Explanation:

<u>Given:</u>

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  • Distance of separation between the plates, d = 1.0 cm = 0.01 m.
  • Minimum value of electric field that produces spark, \rm E=3\times 10^6\ N/C.

When the dimensions of the plate of the capacitor is comparatively much larger than the distance of separation between the plates, then, according to the Gauss' law of electrostatics, the value of the electric field strength in the region between the plates of the capacitor is given by

\rm E=\dfrac{\sigma}{\epsilon_o}.

where,

  • \rm \sigma = surface charge density of the plate of the capacitor = \dfrac qA.
  • \rm q = magnitude of the charge on each of the plate.
  • \rm A = surface area of each of the plate =\rm \pi \times (Radius)^2=\pi \times\left ( \dfrac{D}{2}\right )^2= \pi \times \left ( \dfrac{0.21}{2}\right )^2=3.46\times 10^{-2}\ m^2.
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For the minimum value of electric field that produces spark,

\rm E = \dfrac{q}{A\epsilon_o}\\\Rightarrow q = E\ A\epsilon_o\\=3\times 10^6\times 3.46\times 10^{-2}\times 8.85\times 10^{-12}\\=9.186\times 10^{-7}\ C.

It is the maximum value of the magnitude of charge which can be added up to each of the plates of the capacitor.

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3 years ago
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Explanation:

Given that,

The respective indices of refraction for the blue light and the red light are 1.4636 and 1.4561.

A ray of light consisting of blue light (wavelength 480 nm) and red light (wavelength 670 nm) is incident on a thick piece of glass at 80 degrees.

We need to find the angular separation between the refracted red and refracted blue beams while they are in the glass.

Using Snell's law for red light as :

n_1\sin\theta_1=n_2\sin\theta_2\\\\\theta_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta_2=\sin^{-1}((\dfrac{1}{1.4561})\sin(80))\\\\\theta_2=42.555

Again using Snell's law for blue light as :

n_1\sin\theta_1=n_2\sin\theta'_2\\\\\theta'_2=\sin^{-1}((\dfrac{n_2}{n_1})\sin\theta_1)\\\\\theta'_2=\sin^{-1}((\dfrac{1}{1.4636 })\sin(80))\\\\\theta'_2=42.283

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Answer:

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Explanation:

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U=Q*V

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Potential\ Energy=\frac{Qkq}{r} \\Q=q\\Potential\ Energy=\frac{kq^2}{r}

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q is the charge on electron=-1.6*10^-19 C

r is the distance=3.0*10^{-10}m

For 3 Electrons Potential Energy or work Done is:

Potential\ Energy=3*\frac{kq^2}{r}

Potential\ Energy=3*\frac{(8.99*10^9)(-1.6*10^{-19})^2}{3*10^{-10}}\\Potential\ Energy=2.301*10^{-18} J

Work  done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is 2.301*10^{-18} Joules

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3 years ago
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