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dangina [55]
2 years ago
12

8.4-1 Consider a magnetic field probe consisting of a flat circular loop of wire with radius 10 cm. The probe’s terminals corres

pond to a small gap in the loop. This probe is placed in a uniform magnetic field having magnitude B(t) = B0 sin(2πf t + α), where f = 100 kHz and α is an unknown constant. (By uniform, we mean that the magnetic field has the same magnitude and direction at all points in space.) The orientation of the loop with respect to the magnetic field vector is unknown. The voltage at the terminals is measured for all possible orientations of the probe, and it is found that the maximum voltage is 20 mV peakto-peak. What is B0?
Physics
1 answer:
Vlad1618 [11]2 years ago
6 0

Answer:

B_o = 1.013μT

Explanation:

To find B_o you take into account the formula for the emf:

\epsilon=-\frac{d\Phi_b}{dt}=-\frac{dBAcos\theta}{dt}=-Acos\theta\frac{dB}{dt}

where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.

By applying the derivative you obtain:

\epsilon=-Acos\theta (2\pi f) B_ocos(2\pi f t+ \alpha)

when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:

\epsilon=2\pi fAB_o=2\pi (100*10^3Hz)(\pi (0.1m)^2)B_o=19739.20Hzm^2B_o\\\\B_o=\frac{20*10^{-3}V}{19739.20Hzm^2}=1.013*10^{-6}T=1.013\mu T

hence, B_o = 1.013μT

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zhuklara [117]

Answer:

<em>You would use the kinematic formula:</em>

    \Delta y=V_{0y}\times t-g\times t^2/2

Explanation:

The upwards vertical motion is ruled by the equation:

        y=y_0+V_{0y}\times t-g\times t^2/2

Where:

       y \text{ is the position at the time }t:y=0.1m

       y_0\text{ is the initial position: }y_0=0

       t=2s

       g\text{ is the gravitational acceleration: }\approx 9.8m/s^2

       V_{0y}\text{ is the initial vertical velocity}

Naming Δy = y - y₀, the equation becomes:

      \Delta y=V_{0y}\times t-g\times t^2/2

Then, you just need to substitute with Δy = 0.1m, t = 2s, and g = 9.8m/s², ans solve for the intital vertical velocity.

7 0
2 years ago
Read 2 more answers
You are seated in a bus and notice that a hand strap that is hanging from the ceiling hangs away from the vertical in the backwa
Murrr4er [49]

Answer:C

Explanation:

It is given that hand strap moves from the vertical in the backward direction.

The direction of strap depends upon the acceleration of bus i.e. if bus is accelerating in forward direction then strap will move in backward direction and vice-versa.

The reason for moving backwards is due to the psuedo acting on strap which bends the strap in backward direction

angle of inclination is given by \tan \theta =\frac{a}{g}

where a=acceleration of bus

\theta=inclination of strap from vertical

so we cannot conclude anything about the direction of the velocity of the bus

3 0
2 years ago
The density of nuclear matter is about 1018 kg/m3. Given that 1 mL is equal in volume to 1 cm3, what is the density of nuclear m
Sonbull [250]

Answer:

density is 10^{6} Mg/µL

Explanation:

given data

density of nuclear = 10^{18} kg/m³

1 ml = 1 cm³

to find out

density of nuclear matter in Mg/µL

solution

we know here

1 Mg = 1000 kg

so

1 m³ is equal to 10^{6} cm³

and here 1 cm³ is equal to  1 mL

so we can say 1 mL is equal to 10³ µL

so by these we can convert density

density = 10^{18} kg/m³

density = 10^{18} kg/m³ × \frac{10^{-3} }{10^{6} }  Mg/µL

density =  10^{6} Mg/µL

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2 years ago
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In a 1.25-T magnetic field directed vertically upward, a particle having a charge of magnitude 8.50μC and initially moving north
Goshia [24]

Answer:

a) The sign of the charge is positive.

b) The magnetic force on the particle is 0.050 newtons.

Explanation:

The magnetic force F on a moving charge with velocity v passing through a magnetic field B is:

\overrightarrow{F}=q\overrightarrow{v}\times\overrightarrow{B}(1)

a)

Because it is a cross product, we can find the direction of the force using the right-hand rule, that is too the direction of the movement. We have two possibilities here because the velocity vector and magnetic field are perpendicular: the particle deflects towards east or toward west, which depends on the charge of the particle. Note that if you put your right hand fingers, except thumb, pointing towards north (direction of velocity) and later close them in the direction of the magnetic field, if you maintain your thumb perpendicular to this movement it will point towards east (See figure), so that will be de direction of the force if the charge is positive, but if the charge is negative, the direction will be opposite (towards west). So the charge has to be positive to deflects towards east.

b)

Now by 1:

F=qvB\sin\theta=(8.50\times10^{-6})(4750)(1.25)\sin90\simeq\mathbf{0.050\,N}

5 0
2 years ago
You throw a Frisbee of mass m and radius r so that it is spinning about a horizontal axis perpendicular to the plane of the Fris
Usimov [2.4K]

Answer:

Torque τ =w ×0 = 0

Explanation:

We know that the torque is given by the product of the force and perpendicular distance between the force and the axis.

Here the gravity force act at the center and the rotational axis is also passing through the center.

Therefore the perpendicular distance between the force and the rotational axis would be zero.

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Therefore the torque would be zero.

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