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svetlana [45]
2 years ago
7

A circular loop of wire is rotated at constant angular speed about an axis whose direction can be varied. In a region where a un

iform magnetic field points straight down, what must be the orientation of the loop's axis of rotation if the induced emf is to be zero?
Physics
1 answer:
katovenus [111]2 years ago
5 0

Answer:

here the coil must be oriented in such a way that its plane is perpendicular to the magnetic field

Explanation:

As we know by Faraday's law of electromagnetic induction

Rate of change in magnetic flux will induce EMF in the coil

so here we will have

EMF = \frac{d\phi}{dt}

here we know that

\phi = NB.A

now if the magnetic flux will change with time then it will induce EMF in the coil

EMF = N\frac{d}{dt}(B.A)

so here induced EMF will be zero in the coil if the flux linked with the coil will remain constant

so here the coil must be oriented in such a way that its plane is perpendicular to the magnetic field

In such a way when coil will rotate then the flux linked with the coil will remains constant and there will be no induced EMF in it

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Let v1, , vk be vectors, and suppose that a point mass of m1, , mk is located at the tip of each vector. The center of mass for
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Answer:

Explanation:

Center of mass is give as

Xcm = (Σmi•xi) / M

Where i= 1,2,3,4.....

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x is the position of the mass (x, y)

Now,

Given that,

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u2 = (2, 1, −3) (mass 1 kg),

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u3 = (0, 4, 3) (mass 2 kg),

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u4 = (5, 2, 0) (mass 5 kg)

m4 = 5kg and it position x4 = (5,2,0)

Now, applying center of mass formula

Xcm = (Σmi•xi) / M

Xcm = (m1•x1+m2•x2+m3•x3+m4•x4) / (m1+m2+m3+m4)

Xcm = [3(-1, 0, 2) +1(2, 1, -3)+2(0, 4, 3)+ 5(5, 2, 0)]/(3 + 1 + 2 + 5)

Xcm = [(-3, 0, 6)+(2, 1, -3)+(0, 8, 6)+(25, 10, 0)] / 11

Xcm = (-3+2+0+25, 0+1+8+10, 6-3+6+0) / 11

Xcm = (24, 19, 9) / 11

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This is the required center of mass

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AleksAgata [21]

Answer:

Explanation:

Let L be the length of the wire.

velocity of pulse wave v = L / 24.7 x 10⁻³ = 40.48 L  m /s

mass per unit length of the wire m = 14.5 x 10⁻⁶ x 10⁻³ / 2 x 10⁻² kg / m

m = 7.25 x 10⁻⁷ kg / m

Tension in the wire = Mg  , M is mass hanged from lower end.

= .4 x 9.8

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expression for velocity of wave in the wire

v = \sqrt{\frac{T}{m} }    , T is tension in the wire , m is mass per unit length of wire .

40.48 L = \sqrt{\frac{3.92}{7.25\times10^{-7}} }

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Answer:

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