Physics, 08.11.2019 06:31, athenajames1221

Electrons (mass m, charge –e) are accelerated from rest through a potential difference v and are then deflected by a magnetic field that is perpendicular to their velocity. the radius of the resulting electron trajectory is:

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The correct answer was given: j4ckd4ws

$r=\dfrac{1}{B}\sqrt{\dfrac{2Vm}{e}}$

Explanation:

Let m and e are the mass and charge of an electron. It is accelerated from rest through a potential difference V and are then deflected by a magnetic field that is perpendicular to their velocity. Let v is the velocity of the electron. It can be calculated as :

$\dfrac{1}{2}mv^2=eV$

$v=\sqrt{\dfrac{2eV}{m}}$

When the electron enters the magnetic field, the centripetal force is balanced by the magnetic force as :

$\dfrac{mv^2}{r}=evB$

$r=\dfrac{mv}{eB}$

$r=\dfrac{1}{B}\sqrt{\dfrac{2Vm}{e}}$

So, the radius of the resulting electron trajectory is $\dfrac{1}{B}\sqrt{\dfrac{2Vm}{e}}$ . Hence, this is the required solution.

The correct answer was given: AleciaCassidy

(a) 5172 V/m

The electric field strength of a uniform field between two parallel plates is given by:

$E=\frac{V}{d}$

where

V is the potential difference between the plates

d is the distance between the plates

Here we have

V = 3.00 x 10^3 V

d = 0.580 m

So the electric field strength is

$E=\frac{3\cdot 10^3}{0.580}=5172 V/m$

(b) 1.74 x 10^5 m/s

According to the law of conservation of energy, the electric potential energy of the ion is converted into kinetic energy. Therefore we can write:

U=K

$q\Delta V = \frac{1}{2}mv^2$

where

$q= 1e=1.6\cdot 10^{-19}C$ is the charge of the ion

$\Delta V=3\cdot 10^3 V$ is the potential difference

$m=3.17\cdot 10^{-26} kg$ is the mass of the ion

v is the final speed of the ion

Solving for v, we find

$v=\sqrt{\frac{2q\Delta V}{m}}=\sqrt{\frac{2(1.6\cdot 10^{-19})(3000)}{3.17\cdot 10^{-26}}}=1.74\cdot 10^5 m/s$

When the ion moves into the magnetic field region, the magnetic force acting on it acts as centripetal force.

The magnetic force is:

F = qvB

B = 5.19 x 10^−2 T is the magnetic field strength

v is the speed of the ion, found in the previous part

While the centripetal force is

$F=m\frac{v^2}{r}$

where

r is the radius of the circular path

So we can write

$qvB = m\frac{v^2}{r}$

and solving for the radius, we find

$r=\frac{mv}{qB}=\frac{(3.17\cdot 10^{-26})(1.74\cdot 10^5)}{(1.6\cdot 10^{-19})(5.19\cdot 10^{-2})}=0.664 m$

The correct answer was given: Izayoi

Alpha and beta radiation can also be deflected by magnetic fields . Just as with electric fields, gamma radiation is not deflected by magnetic fields.

The correct answer was given: 2sally2

Gamma is NOT deflected by a magnetic field.

The correct answer was given: cadeedmiston

Gamma radiation isn't deflected

Explanation:

Hi, Rutherford characterized three types of radioactivity: Alfa, Beta and Gamma.

The principal difference is that the first two are charged particules (+ or -) but the gamma radiation comes from neutral particules.

Magnetic fields have the ability to interact with charged particules (like ions) or with ionizable elements (like the ferritic ones).

Being that said, the gamma radiation won't be deflected by the magnetic field because of its neutrality.

The correct answer was given: bobiscool3698

R = 0.584m

Explanation:

Given B = 537G = 537×10-⁴T = 0.054 T

Mass, m = 3.98×10-²⁶ kg, V = 2.00kV = 2000V. q = 1.6×10-¹⁹ C

The radius R of curvature is given by

R = mv/qB

We don't know the value for v the velocity, but we can get the expression for it from energy concepts.

The electric potential energy of the field is equal to the kinetic energy of the ion

qV = 1/2mv²

v² = 2qV/m = 2 ×1.6×10-¹⁹×2000 /(3.98×10-²⁶)

v² = 1.608×10¹⁰

v = sqrt(1.608×10¹⁰)= 126809m/s

So R = 3.98×10-²⁶ × 126809/(1.6×10-¹⁹ × 0.054)

R = 0.584 m

The correct answer was given: gabbycasey102

The radius of curvature of the ion's orbit is 0.59 meters

Explanation:

Given that,

Mass of the 24 Mg ion, $m=3.983\times 10^{-26}\ kg$

Potential difference, V = 3 kV

Magnetic field, B = 526 G

Charge on single ionized ion, $q=1.6\times 10^{-19}\ C$

The radius of the the path traveled by the charge is circular. Its radius is given by :

$r=\dfrac{mv}{Bq}$

v is speed of particle.

v can be calculated using conservation of energy as :

$\dfrac{1}{2}mv^2=qV\\\\v=\sqrt{\dfrac{2qV}{m}} \\\\v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 2\times 10^3}{3.983 \times 10^{-26}}} \\\\v=1.26\times 10^5\ m/s$

Radius,

$r=\dfrac{3.983 \times 10^{-26}\times 1.26\times 10^5}{0.0526\times 1.6\times 10^{-19}}\\\\r=0.59\ m$

So, the radius of curvature of the ion's orbit is 0.59 meters.

The correct answer was given: preservations

they have a negligible mass and they also have a negative charge so they have an association with the magnetic field. Do you remember polar? So electrons are negative and they attract the nucleus which is overall postivie (since protons are positive) so they act as magnetic since they both have an attraction

The correct answer was given: savannahvargas512

GAMMA PARTICLES DOES NO BECOME DEFLECED

The correct answer was given: cass9980

There are four key characteristics of the magnetic force on a moving charge.

First, Its magnitude is proportional to the magnitude of the charge.
Second, the magnitude of the force is also proportional
Third, the magnetic force depends on the particle’s
velocity.
Fourth, we find by experiment that the magnetic force $\vec{F}$ does not have the same direction as the magnetic field $\vec{F}$ but instead is always perpendicular to both $\vec{B}$ and the velocity $\vec{v}$

So, in a mathematical language this is given by:

$\vec{F}=q\vec{v} \times \vec{B}$

So, this is a cross product. Therefore, applying the Right-hand rule:

The answer is:

D. to the left of its velocity.

Explaining this in other words:

"The force $\vec{F}$ is directed into the plane of the paper, so this is the direction of the proton"
Aproton is moving horizontally at 7.9 x 10^5 m/s. it passes through a vertical magnetic field that p