PHYSICS COLLEGE

An 800-kHz sinusoidal radio signal is detected at a point 6.6 km from the transmitter tower. The electric field amplitude of the signal at that point is 0.780 V/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. What is the amplitude of the magnetic field of the signal at that point

Answers

Answer 1

Answer:

Answer:

0.0000000026 T

Explanation:

$E_0$ = Maximum electric field strength = 0.78 V/m

$B_0$ = Maximum magnetic field strength

c = Speed of light = $3* 10^8\ m/s$

Relation between amplitudes of electric and magnetic fields is given by

$E_0=B_0c\n\Rightarrow B_0=(E_0)/(c)\n\Rightarrow B_0=(0.78)/(3* 10^8)\n\Rightarrow B_0=0.0000000026\ T$

The amplitude of the magnetic field is 0.0000000026 T

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Two large parallel metal plates are 1.6 cm apart and have charges of equal magnitude but opposite signs on their facing surfaces. Take the potential of the negative plate to be zero. If the potential halfway between the plates is then +3.8 V, what is the electric field in the region between the plates?
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A proton moves perpendicular to a uniform magnetic field B with arrow at a speed of 2.20 107 m/s and experiences an acceleration of 1.90 1013 m/s2 in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field.

Answers

Answer:

The magnitude and direction of the magnetic field is 0.009014 T in the negative y direction.

Explanation:

Given that,

Speed $v = 2.20*10^7\ m/s$

Acceleration $a=1.90*10^(13)\ m/s^2$

We need to calculate the magnetic field

Using formula of magnetic field

$F=qvB$ ....(I)

Using newton's second law

$F= ma$ ....(II)

From equation (I) and (II)

$ma=qvB$

Put the value into the formula

$1.90*10^(13)*1.67*10^(-27)=1.6*10^(-19)*2.20*10^(7)*B$

$3.173*10^(-14)=1.6*10^(-19)*2.20*10^(7)*B$

$B=(3.173*10^(-14))/(1.6*10^(-19)*2.20*10^(7))$

$B=0.009014\ T$

We need to calculate the direction of the field

Using the right hand rule, point the right hand fingers along the velocity which is in the positive z direction.

Now, if we curl the fingers along the direction of magnetic field that is in the negative y direction, then the thumb will point in the positive x direction.

Hence, The magnitude and direction of the magnetic field is 0.009014 T in the negative y direction.

2. Annealed low-carbon steel has a flow curve with strength coefficient of 75000 psi and strain-hardening exponent of 0.25. A tensile test specimen with a gauge length of 2 in. is stretched to a length of 3.3 in. Determine the flow stress and the average flow stress that the metal experienced during this deformation.

Answers

Answer:

Flow stress= 9390Psi

Average flow stress= 4173.33Psi

Explanation:

Given:

Strength Coefficient = 75000psi

Strain hardening Exponent = 0.25

Gauge length = 2inches

Stretch length = 3.3 inches

Flow stress,Yf = 75000 × ln(3.3/2) × 0.25

Yf = 75000× ln(1.65) × 0.25

Yf = 75000× 0.5008 × 0.25

Flow stress = 9390Psi

Average flow stress = 75000× 0.5008 × (0.25/2.25)

Average flow stress= 4173.33psi

Two parallel wires carry currents in the same direction. If the currents in the wires are 1A and 4A and the wires are 5 m apart. Calculate the force between the 2 wires. Is it attractive or repulsive? Calculate the magnetic field midway between the wires.

Answers

Answer:

$1.6* 10^(-7)$ N

$2.4* 10^(-7)$ N

Explanation:

$i_(1)$ = 1 A

$i_(2)$ = 4 A

$r$ = distance between the two wire = 5 m

$F$ = Force per unit length acting between the two wires

Force per unit length acting between the two wires is given as

$F = (\mu _(o))/(4\pi )(2i_(1)i_(2))/(r)$

$F = (10^(-7))(2(1)(4))/(5)$

$F = 1.6* 10^(-7)$ N

$r'}$ = distance of each wire from the midpoint = 2.5 m

Magnetic field midway between the two wires is given as

$B = (\mu _(o))/(4\pi ) \left \left ( (2i_(2))/(r') \right - (2i_(1))/(r') \right \right ))$

$B = (10^(-7)) \left \left ( (2(4))/(2.5) \right - (2(1))/(2.5) \right \right ))$

$B = 2.4* 10^(-7)$

A charge of uniform volume density (40 nC/m3) fills a cube with 8.0-cm edges. What is the total electric flux through the surface of this cube?

Answers

Answer:

The flux through the surface of the cube is $2.314\ Nm^(2)/C$

Solution:

As per the question:

Edge of the cube, a = 8.0 cm = $8.0* 10^(- 2)\ m$

Volume Charge density, $\rho_(v) = 40 nC/m^(3) = 40* {- 9}\ C/m^(3)$

Now,

To calculate the electric flux:

$\phi = (q)/(\epsilon_(o))$ (1)

where

$\phi$ = electric flux

$\epsilon_(o) = 8.85* 10^(- 12)\ F/m$ = permittivity of free space

Volume Charge density for the given case is given by the formula:

$\rho_(v) = (Total\ charge, q)/(Volume of cube, V)$ (2)

Volume of cube, $V = a^(3)$

Thus

$V = (8.0* 10^(- 2))^(3) = 5.12* 10^(- 4)\ m^(3)$

Thus from eqn (2), the total charge is given by:

$q = \rho_(v)V = 40* {- 9}* 5.12* 10^(- 4)$

$q = 2.048* 10^(-11)\ F = 20.48\ pF$

Now, substitute the value of 'q' in eqn (1):

$\phi = (2.048* 10^(-11))/(8.85* 10^(- 12)) = 2.314\ Nm^(2)/C$

A Hall-effect probe to measure magnetic field strengths needs to be calibrated in a known magnetic field. Although it is not easy to do, magnetic fields can be precisely measured by measuring the cyclotron frequency of protons. A testing laboratory adjusts a magnetic field until the proton's cyclotron frequency is 9.70 MHz . At this field strength, the Hall voltage on the probe is 0.549 mV when the current through the probe is 0.146 mA . Later, when an unknown magnetic field is measured, the Hall voltage at the same current is 1.735 mV .A) What is the strength of this magnetic field?

Answers

Answer:

The value of the magnetic field is 2.01 T when Hall voltage is 1.735 mV

Explanation:

The frequency of the cyclotron can help us find the magnitude of the magnetic field, thus then we can compare the effect of increasing Hall voltage on the probe.

Magnetic field magnitude at initial Hall voltage.

The cyclotron frequency can be written in terms of the magnetic field magnitude as follows

$f = \cfrac{qB}{2\pi m}$

Solving for the magnetic field.

$B = \cfrac{2\pi mf}q$

Thus we can replace the given information but in Standard units, also remembering that the mass of a proton is $m_p=1.67 * 10^(-27) kg$ and its charge is $q_p=1.6 * 10^(-19) C$ .

So we get

$B = \cfrac{2\pi * 1.67 * 10^(-27) kg * 9.7 * 10^6 Hz}{1.6 * 10^(-19)C}$

$B =0.636 T$

We have found the initial magnetic field magnitude of 0.636 T

Magnetic field magnitude at increased Hall voltage.

The relation given by Hall voltage with the magnetic field is:

$V_H =\cfrac{R_HI}t B$

Thus if we keep the same current we can write for both cases:

$V_(H1) =\cfrac{R_HI}t B_1\nV_(H2) =\cfrac{R_HI}t B_2$

Thus we can divide the equations by each other to get

$\cfrac{V_(H1) }{V_(H2)}=\cfrac{\cfrac{R_HI}t B_1}{\cfrac{R_HI}t B_2}$

Simplifying

$\cfrac{V_(H1) }{V_(H2)}=\cfrac{ B_1}{ B_2}$

And we can solve for $B_2$

$B_2 =B_1 \cfrac{V_(H2)}{V_(H1)}$

Replacing the given information we get

$B_2= 0.636 T * \left(\cfrac{1.735 mV}{0.549 mV} \right)$

We get

$\boxed{B=2.01\, T}$

Thus when the Hall voltage is 1.735 mV the magnetic field magnitude is 2.01 T

A twirlers baton is 0.76 m long and spins around its center. The end of the baton has a centripetal acceleration of 47.8 m/s2?A 0.31 s
B 0.56 s
C 4.3s
D 70s

Answers

B 0.56 s is the time period of a twirlers baton.

What is Centripetal Acceleration?

Centripetal acceleration is defined as the property of the motion of an object which traversing a circular path.

Any object that is moving in a circle and has an acceleration vector pointed towards the center of that circle is known as Centripetal acceleration.

The centripetal acceleration is given by:

a = 4π²R/T²

Given values are:

a = 47.8 m/s²

D = 0.76 m so , R = 0.76/2 = 0.38m

Using this formula,

47.8*T² = 4π² x0.38

T² = $(4*3.14^2*0.38)/(47.8)$

T = 0.56 s

Therefore,

A twirlers baton is 0.76 m long and spins around its center. The end of the baton has a centripetal acceleration of 47.8 m/s2 which have time period of 0.56 s.

Learn more about Centripetal acceleration here:

brainly.com/question/14465119

#SPJ5

Answer:

C. 4.3 seconds

Explanation:

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