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pav-90 [236]
3 years ago
9

If a circular loop of wire of radius 19.7 cm is located in a region where the spatially uniform magnetic field perpendicular to

the plane of the loop is changing at a rate of +2.0 ✕ 10−3 T/s, find the value of the induced EMF in this loop due to this changing magnetic field.
Physics
1 answer:
Alisiya [41]3 years ago
6 0

Answer:

Magnitude of induced emf will be 2.437\times 10^{-3}volt

Explanation:

We have given radius of the circular loop r = 19.7 cm = 0.197 m

So area of the circular loop A=\pi r^2=3.14\times 0.197^2=0.1218m^2

It is given that magnetic field is changing as 2\times 10^{-3}T/sec

Emf induced in the circular loop is equal to e=\frac{-d\Phi }{dt}=-A\frac{dB}{dt}

So emf induced will be equal to e=\frac{-d\Phi }{dt}=-0.1218\times2\times 10^{-3}=2.437\times 10^{-3}volt

So magnitude of induced emf will be 2.437\times 10^{-3}volt

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Explanation:

Given that,

Mass of disk = 1.2 kg

Radius = 0.07 m

Radius of rod = 0.11 m

Mass of small disk = 0.5 kg

Force = 29 N

Time t = 0.022 s

\theta=0.023\ m

Distance d= 0.039 m

(I). We need to calculate the speed of the apparatus

Using work energy theorem

W=\Delta K.E

Fd=\dfrac{1}{2}mv^2

v=\sqrt{\dfrac{2Fd}{M+4m}}

Where, m = total mass

v = velocity

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d = distance

Put the value into the formula

v=\sqrt{\dfrac{2\times29\times0.039}{1.2+4\times0.5}}

v=0.840\ m/s

(b). We need to calculate the angular speed of the apparatus

Using formula of torque

\tau=I\alpha

F\times=(\dfrac{1}{2}MR^2+4mb^2)\alpha

29\times0.07=(\dfrac{1}{2}\times1.2\times0.07^2+4\times0.5\times0.11^2)\alpha

\alpha=\dfrac{29\times0.07}{0.02714}

\alpha=74.79\ rad/s^2

We need to calculate the angular speed of the apparatus

Using equation of angular motion

\omega=\omega_{0}+\alpha t

Put the value into the formula

\omega=0+74.79\times0.022

\omega=1.645\ rad/s

(c).  We need to calculate the angular speed of the apparatus

Using equation of angular motion

\omega_{0}^2=\omega^2+2\alpha t

Put the value into the formula

\omega_{0}^2=1.645^2+2\times74.79\times0.022

\omega=2.44\ rad/s

Hence, This is required equation.

7 0
3 years ago
What type of circuit is illustrated?A)closed series circuit
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I think it’s B open series circuit but I’m not sure
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3 years ago
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A ball is thrown upwards at an unknown speed. in a time of
alexandr402 [8]

Answer:

a)  Initial speed of the ball = 14.45 m/s

b) At height 6 m speed of ball = 9.55 m/s

c) Maximum height reached = 10.65 m

Explanation:

a)  We have equation of motion s=ut+\frac{1}{2} at^2, where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.

s = 6 m, t = 0.5 seconds, a = acceleration due to gravity value = -9.8m/s^2

 Substituting

    6=u*0.5-\frac{1}{2} *9.8*0.5^2\\ \\ u=14.45m/s

 Initial speed of the ball = 14.45 m/s

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        v^2=14.45^2-2*9.8*6\\ \\ v=9.55m/s

  So at height 6 m speed of ball = 9.55 m/s

c) We have equation of motion v^2=u^2+2as, where v is the final velocity

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   Substituting

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A plate carries a charge of 3.8 UC, while a rod carries a charge of 1.9 C. How many electrons must be transferred from the plate
frosja888 [35]

Answer:

N_{electrons}=Q_{transfered}/q_{electron}=5.94*10^{18}electrons

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The plate and the rod must have Q_{total}/2\\. So the charge transferred from the plate to the rod is:

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3 years ago
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