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Maslowich
3 years ago
7

A loop of wire of area A is tipped at an angle eto uniform magnetic field B. The maximum flux occurs for an angle A=0. What angl

e 0 will give a flux that is of this maximum value? A. 0=30° B. 0=45 c. 0=60 d.0=90
Physics
1 answer:
trasher [3.6K]3 years ago
8 0

Answer:

We have that θ is the angle that makes the normal of the loop and the direction of the magnetic field.

This means that if θ = 0°, the plane in which the loop of wire lies is totally perpendicular to the magnetic field. (or the face of the loop points in the same direction than the field)

Now, the magnetic flux can be calculated as:

\int\limits^} \,B.dA

Where dA is the differential of area, and we have a dot product, then we have:

B.dA = B*Acos(θ)ds

ds is a differential of surface.

Now, the flux will be maximum when cos(θ) is also a maximum.

And the maximum of the cosine is:

Cos(0°) = 1.

decreasing until cos(90°) = 0

Now, the options given are:

30°, 45°, 60° and 90°.

Then in this range, the maximum flux will occur at the angle closer to 0°, then the correct option is θ = 30°

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zubka84 [21]

The force needed to accelerate an elevator upward at a rate of 2 m / s^{2} is 2000 N or 2 kN.

<u>Explanation: </u>

As per Newton's second law of motion, an object's acceleration is directly proportional to the external unbalanced force acting on it and inversely proportional to the mass of the object.

As the object given here is an elevator with mass 1000 kg and the acceleration is given as 2 m / s^{2}, the force needed to accelerate it can be obtained by taking the product of mass and acceleration.

                  \text {Force}=\text {Mass} \times \text {Acceleration}

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3 years ago
The parallel plates in a capacitor, with a plate area of 7.90 cm2 and an air-filled separation of 2.70 mm, are charged by a 7.90
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Answer:

A) 26V

Explanation:

(a) the potential difference between the plates

Initial capacitance can be calculated using below expresion

C1= A ε0/ d1

Where d1= distance between = 2.70 mm= 2.70× 10^-3 m

ε0= permittivity of space= 8.85× 10^-12 Fm^-1

A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2

If we substitute the values we

C1= A ε0/ d1

=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3

C1=2.589 ×10^-12 F= 2.59 pF

Initial charge can be determined using below expresion

q1= C1 × V1

V1=2.589 ×10^-12 F

V1= voltage=7.90 V

If we substitute we have

q1= 2.589 ×10^-12 × 7.90

q1= 20.45×10^-12C

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Final capacitance can be calculated as

C2= A ε0/ d2

d2=8.80 mm= /8.80× 10^-3

7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3

C1=0.794 ×10^-12 F= 0.794 pF

Final charge= initial charge

q2=q1 (since the battery is disconnected)

q2=q1= 20.45 pC

Final potential difference

V2= q/C2

= 20.45/0.794

= 26V

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