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Wittaler [7]
3 years ago
7

A coil of wire with 60 turns lies on the plane of the page, and a uniform magnetic field points into the page with a strength of

1.10 T. The coil has an initial area of 0.890 m2 and is then elongated so that its area decreases to 0.270 m2 in 0.80 s. What is the magnitude of the average induced emf in the coil while its area is changing? 59.334 Incorrect: Your answer is incorrect. V
Physics
1 answer:
baherus [9]3 years ago
4 0

Answer:

0.85V

Explanation:

The emf is calculated by using the Lenz's Law

\epsilon=-\frac{\Delta\Phi_B}{\Delta t}

But for this case we have that the magnetic field is constant. Hence we have

\epsilon=-\frac{B(A_{2}-A_{1})}{t_2-t_1}=-\frac{(1.10T)(0.270m^{2}-0.89m^{2})}{0.8s}\\\\\epsilon=0.85V

where we have taken that the intial time is t1=0

I hope this is useful for you

regards

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lesya692 [45]

\\ \bull\sf\dashrightarrow F=ma

\\ \bull\sf\dashrightarrow a=\dfrac{F}{m}

Now

\\ \bull\sf\dashrightarrow a\propto\dfrac{1}{m}

\\ \bull\sf\dashrightarrow a\propto F

  • Lower mass=Higher acceleration
  • Lower Force=Lower Acceleration

Option B has lowest mass and highest force hence its correct

8 0
2 years ago
A proton, starting from rest, accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040
Minchanka [31]

Answer:

r = 0.11 m

Explanation:

The radius of the proton's resulting orbit can be calculated equaling the force centripetal (Fc) with the Lorentz force (F_{B}), as follows:

F_{c} = F_{B} \rightarrow \frac{m*v^{2}}{r} = qvB (1)

<u>Where:</u>

<em>m: is the proton's mass =  1.67*10⁻²⁷ kg</em>

<em>v: is the proton's velocity</em>

<em>r: is the radius of the proton's orbit</em>

<em>q: is the proton charge = 1.6*10⁻¹⁹ C</em>

<em>B: is the magnetic field = 0.040 T </em>

Solving equation (1) for r, we have:

r = \frac{mv}{qB}   (2)

By conservation of energy, we can find the velocity of the proton:

K = U \rightarrow \frac{1}{2}mv^{2} = q*\Delta V   (3)

<u>Where:</u>

<em>K: is kinetic energy</em>

<em>U: is electrostatic potential energy</em>

<em>ΔV: is the potential difference = 1.0 kV </em>

Solving equation (3) for v, we have:

v = \sqrt{\frac{2q\Dela V}{m}} = \sqrt{\frac{2*1.6 \cdot 10^{-19} C*1.0 \cdot 10^{3} V}{1.67 \cdot 10^{-27} kg}} = 4.38 \cdot 10^{5} m/s  

Now, by introducing v into equation (2), we can find the radius of the proton's resulting orbit:

r = \frac{mv}{qB} = \frac{1.67 \cdot 10^{-27} kg*4.38 \cdot 10^{5} m/s}{1.6 \cdot 10^{-19} C*0.040 T} = 0.11 m

Therefore, the radius of the proton's resulting orbit is 0.11 m.

I hope it helps you!  

5 0
3 years ago
What force is needed to accelerate an object 5 m/s2 if the object has a mass of 10 kg?
Helga [31]
We know, F = m * a
F = 10 * 5
F = 50 N

In short, Your Answer would be 50 Newtons

Hope this helps!
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Find the radius using the centripetal acceleration formula: acceleration = 6.6 x 10^6 m/s^2 velocity = 2,000 rev/s
aksik [14]

the radius is 231

Explanation:

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kolbaska11 [484]

Answer:

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