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 1

Answer:

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

Related Questions

If the population of penguins increased, then this would have a direct effect on the populations of?
Light of wavelength 608.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 88.5 cm from the slit. The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm. What is the width of the slit?
A baseball player standing on a platform throws a baseball out over a level playing field. The ball is released from a point 3.50 m above the field with an initial speed of 14.3 m/s at an upward angle of 27.0 degrees above the horizontal. What horizontal distance will the ball travel before hitting the ground?
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the
Dr. John Paul Stapp was a U.S. Air Force officer who studied the effects of extreme acceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.2 s and was brought jarringly back to rest in only 1 s. Calculate his (a) magnitude of acceleration in his direction of motion and (b) magnitude of acceleration opposite to his direction of motion. Express each in multiples of g (9.80 m/s2) by taking its ratio to the acceleration of gravity. g g

wo balls have the same mass of 5.00 kg. Suppose that these two balls are attached to a rigid massless rod of length 2L, where L = 0.550 m. One is attached at one end of the rod and the other at the middle of the rod. If the rod is held by the open end and rotates in a circular motion with angular speed of 45.6 revolutions per second,

Answers

Answer:

$T_1 =677224.40\ N$

Explanation:

given,

mass of the both ball = 5 Kg

length of rod = 2 L

where L = 0.55 m

angular speed = 45.6 rev/s

ω = 45.6 x 2 π

ω = 286.51 rad/s

v₁ = r₁ ω₁

v₁ =0.55 x 286.51 = 157.58 m/s

v₂ = r₂ ω₂

v₂ = 1.10 x 286.51 = 315.161 m/s

finding tension on the first half of the rod

r₁ = 0.55 r₂ = 2 x r₁ = 1.10

$T_1 = m ((v_1^2)/(r_1)+(v_2^2)/(r_2))$

$T_1 = 5 ((157.58^2)/(0.55_1)+(315.161^2)/(1.1))$

$T_1 =677224.40\ N$

Tell uses of cancave mirror and convex mirror.

Answers

Answer:

Uses of concave mirror:

Shaving mirrors.

Head mirrors.

Ophthalmoscope.

Astronomical telescopes.

Headlights.

Solar furnaces.

Uses of convex mirror:

Convex mirrors always form images that are upright, virtual, and smaller than the actual object. They are commonly used as rear and side view mirrors in cars and as security mirrors in public buildings because they allow you to see a wider view than flat or concave mirrors.

please give me full points.

You release a block from the top of a long, slippery inclined plane of length l that makes an angle θ with the horizontal. The magnitude of the block's acceleration is gsin(θ).1. For an x axis pointing down the incline and having its origin at the release position, derive an expression for the potential energy of the block-Earth system as a function of x. Suppose that the gravitational potential energy is measured relative to the ground at the bottom of the incline, UG(x=l)=0.Express your answer in terms of g and the variables m, l, x, and θ.(U^G=?)2. Use the expression you derived in the previous part to determine the speed of the block at the bottom of the incline. (Vx,f=?)Express your answer in terms of g and the variables m, l, and θ.

Answers

Answer:

UG (x) = m*g*x*sin(Q)

Vx,f (x)= sqrt (2*g*x*sin(Q))

Explanation:

Given:

- The length of the friction less surface L

- The angle Q is made with horizontal

- UG ( x = L ) = 0

- UK ( x = 0) = 0

Find:

derive an expression for the potential energy of the block-Earth system as a function of x.

determine the speed of the block at the bottom of the incline.

Solution:

- We know that the gravitational potential of an object relative to datum is given by:

UG = m*g*y

Where,

m is the mass of the object

g is the gravitational acceleration constant

y is the vertical distance from datum to the current position.

- We will consider a right angle triangle with hypotenuse x and angle Q with the base and y as the height. The relation between each variable can be given according to Pythagoras theorem as follows:

y = x*sin(Q)

- Substitute the above relationship in the expression for UG as follows:

UG = m*g*x*sin(Q)

- To formulate an expression of velocity at the bottom we can use an energy balance or law of conservation of energy on the block:

UG = UK

- Where UK is kinetic energy given by:

UK = 0.5*m*Vx,f^2

Where Vx,f is the final velocity of the object @ x:

m*g*x*sin(Q) = 0.5*m*Vx,f^2

-Simplify and solve for Vx,f:

Vx,f^2 = 2*g*x*sin(Q)

Hence, Velocity is given by:

Vx,f = sqrt (2*g*x*sin(Q))

A 0.157kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.850m/s . It has a head-on collision with a 0.306kg glider that is moving to the left with a speed of 2.26m/s . Suppose the collision is elastic.Find the magnitude of the final velocity of the 0.157kg glider.Find the magnitude of the final velocity of the 0.306kg glider.

Answers

Answer:

V1 = -3.260 m/s, V2 = 1.303 m/s

Explanation:

Let mass of the left glider m1 = 0.157 kg and velocity v1 = 0.850 m/s

mass of the right glider m2 = 0.306 Kg and v2 = -2.26 m/s (-ve sign mean it is opposite to direction of left glider)

To Find: Final Velocity of Left Glider is V1=? m/s and Velocity of right Glider is V2 =? m/s (After Collision)

from law of conservation of momentum and energy we deduce a formula:

V1 = (m1-m2) v1 /(m1+m2) + 2 m2 v2/(m1+m2)

V1 = (0.157 kg - 0.306 Kg) × 0.850 m/s / (0.157 kg + 0.306 Kg) + 2 ×0.306 kg × -2.26 m/s / (0.157 kg + 0.306 Kg)

V1 = -0.273 -2.987

V1 = -3.260 m/s

and V2 Formula

V2 = (m2-m1) v2/(m1+m2) + 2 m1 v1/(m1+m2)

V2 = (0.157 kg - 0.306 Kg) × -2.26 m/s / (0.157 kg + 0.306 Kg) + 2 ×0.157 kg × 0.850 m/s / (0.157 kg + 0.306 Kg)

V2 = 0.727 + 0.576

V2 = 1.303 m/s

-0.149, 0.463

Tarzan, whose mass is 96 kg, is hanging at rest from a tree limb. Then he lets go and falls to the ground. Just before he lets go, his center of mass is at a height 2.1 m above the ground and the bottom of his dangling feet are at a height 1.3 above the ground. When he first hits the ground he has dropped a distance 1.3, so his center of mass is (2.1 - 1.3) above the ground. Then his knees bend and he ends up at rest in a crouched position with his center of mass a height 0.4 above the ground.Consider the point particle system. What is the speed v at the instant just before Tarzan's feet touch the ground?

Answers

Answer:

5.05 m/s

Explanation:

The distance from the bottom of his feet to his center of mass is (when is hanging at rest) is 2.1 - 1.3 = 0.8 m. Assume he keeps the posture, as soon as his feet touches the ground, his center of mass is 0.8 m above the ground. This would mean that he has traveled a distance of 2.1 - 0.8 = 1.3 m vertically. Using the law of energy conservation for potential and kinetic energy, also let the ground be ground 0 for potential energy, we have the following mechanical conservation energy:

$mgH = mgh + mv^2/2$

Since he was hanging at rest, his initial kinetic energy at H = 2.1m must be 0. Let g = 9.81m/s2 and m be his mass, we can calculate for his velocity v at h = 0.8 m. First start by dividing both sides by m

$gH = gh + v^2/2$

$v^2 = 2g(H - h)$

$v^2 = 2*9.81(2.1 - 0.8) = 25.506$

$v = √(25.506) = 5.05 m/s$

Calculate the ionization potential for C+5 ( 5 electrons removed for the C atom) and in addition compute the wavelength of the transition from n=3 to n= 2.