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ololo11 [35]
2 years ago
14

A rectangular conducting loop of wire is approximately half-way into a magnetic field B (out of the page) and is free to move. S

uppose the magnetic field B begins to decrease rapidly in strength
Requried:
What happens to the loop?

1. The loop is pushed to the left, toward the magnetic field.
2. The loop doesn’t move.
3. The loop is pushed downward, towards the bottom of the page.
4. The loop will rotate.
5. The loop is pushed upward, towards the top of the page.
6. The loop is pushed to the right, away from the magnetic field
Physics
1 answer:
Black_prince [1.1K]2 years ago
8 0

Answer:

. The loop is pushed to the right, away from the magnetic field

Explanation

This decrease in magnetic strength causes an opposing force that pushes the loop away from the field

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Alkali metals _____.
Lyrx [107]
I think the answer is C
7 0
2 years ago
Two objects are 5 kg and 10 kg respectively, and they're 10 m apart. if the distance between them is increased to 20 m, what hap
Sati [7]
I think it's OPTION A, IT DECREASES.
6 0
2 years ago
The magnitude of a force vector is 89.6 newtons (N). The x component of this vector is directed along the +x axis and has a magn
insens350 [35]

Answer:

(a) θ = 33.86°

(b) Ay = 49.92 N

Explanation:

You have that the magnitude of a vector is A = 89.6 N

The x component of such a vector is Ax = 74.4 N

(a) To find the angle between the vector and the x axis you use the following formula for the calculation of the x component of a vector:

A_x=Acos\theta       (1)

Ax: x component of vector A

A: magnitude of vector A

θ: angle between vector A and the x axis

You solve the equation (1) for θ, by using the inverse of cosine function:

\theta=cos^{-1}(\frac{A_x}{A})=cos{-1}(\frac{74.4N}{89.6N})\\\\\theta=33.86\°

the angle between the A vector and the x axis is 33.86°

(b) The y component of the vector is given by:

A_y=Asin\theta\\\\A_y=(89.6N)sin(33.86\°)=49.92N

the y comonent of the vecor is Ay = 49.92 N

3 0
2 years ago
A loop of wire is in a magnetic field such that its axis is parallel with the field direction. Which of the following would resu
ahrayia [7]

Answer:

All the given options will result in an induced emf in the loop.

Explanation:

The induced emf in a conductor is directly proportional to the rate of change of flux.

emf = -\frac{d \phi}{dt} \\\\where;\\\\\phi \ is \ magnetic \ flux\\\\\phi = BA\ cos \theta

where;

A is the area of the loop

B is the strength of the magnetic field

θ is the angle between the loop and the magnetic field

<em>Considering option </em><em>A</em>, moving the loop outside the magnetic field will change the strength of the magnetic field and consequently result in an induced emf.

<em>Considering option </em><em>B</em>, a change in diameter of the loop, will cause a change in the magnetic flux and in turn result in an induced emf.

Option C has a similar effect with option A, thus both will result in an induced emf.

Finally, <em>considering option</em> D, spinning the loop such that its axis does not consistently line up with the magnetic field direction will<em> </em>change the angle<em> </em>between the loop and the magnetic field. This effect will also result in an induced emf.

Therefore, all the given options will result in an induced emf in the loop.

4 0
3 years ago
A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V. What are (a) the total energy stored i
Debora [2.8K]

Answer:

(A) Total energy will be equal to 0.044\times 10^{-5}J

(b) Energy density will be equal to 0.0175J/m^3

Explanation:

We have given diameter of the plate d = 2 cm = 0.02 m

So area of the plate A=\pi r^2=3.14\times 0.02^2=0.001256m^2

Distance between the plates d = 0.50 mm = 0.50\times 10^{-3}m

Permitivity of free space \epsilon _0=8.85\times 10^{-12}F/m

Potential difference V =200 volt

Capacitance between the plate is equal to C=\frac{\epsilon _0A}{d}=\frac{8.85\times 10^{-12}\times 0.001256}{0.50\times 10^{-3}}=0.022\times 10^{-9}F

(a) Total energy stored in the capacitor is equal to

E=\frac{1}{2}CV^2

E=\frac{1}{2}\times 0.022\times 10^{-9}\times 200^2=0.044\times 10^{-5}J

(b) Volume will be equal to V=Ad, here A is area and d is distance between plates

V=0.001256\times 0.02=2.512\times 10^{-5}m^3

So energy density =\frac{Energy}{volume}=\frac{0.044\times 10^{-5}}{2.512\times 10^{-5}}=0.0175J/m^3

7 0
3 years ago
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