Question

A) A mass spectrometer has a velocity selector that allows ions traveling at only one speed to pass with no deflection through slits at the ends. While moving through the velocity selector, the ions pass through a 60,000-N/C E? field and a 0.0500-T B? field. The quantities v?, E?, and B? are mutually perpendicular. Determine the speed of the ions that are not deflected.B) After leaving the velocity selector, the ions continue to move in the 0.0500-T magnetic field. Determine the radius of curvature of a singly charged lithium ion, whose mass is 1.16

Answer 1

Answer:

(A). The speed of the ions is $1.2\times10^{6}\ m/s$

(B). The radius of curvature of a singly charged lithium ion is $2.0\times10^{6}\ m$

Explanation:

Given that,

Electric field = 60000 N/C

Magnetic field = 0.0500 T

(A). We need to calculate the velocity

For no deflection

$F_{E}=F_{B}$

$Eq=Bqv$

$v = \dfrac{E}{B}$

$v=\dfrac{60000}{0.0500}$

$v=1.2\times10^{6}\ m/s$

(B). We need to calculate the radius

Using magnetic force balance by centripetal force

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

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

Put the value into the formula

$r=\dfrac{1.16\times10^{-26}\times(1.2\times10^{6})^2}{0.0500\times1.6\times10^{-19}}$

$r=2.0\times10^{6}\ m$

Hence, (A). The speed of the ions is $1.2\times10^{6}\ m/s$

(B). The radius of curvature of a singly charged lithium ion is $2.0\times10^{6}\ m$

Answer 2

(A)The speed of the ions is 1.2×10⁶ m/s

(B)The radius of curvature of a singly charged lithium-ion is 2.0×10⁶ m.

What is a magnetic field?

It is the type of field where the magnetic force is obtained. With the help of a magnetic field. The magnetic force is obtained it is the field felt around a moving electric charge.

The given electric field is E = 60000 N/C.

The given Magnetic field is B =  0.0500 T.

(A) Velocity of ion =?

In the no deflection condition,

Electric field = Magnetic field

$\rm{ F_E= F_B \\\\Eq=Bqv}$

$\rm v = \frac{E}{B} \\\\\rm v = \frac{60,000}{0.0500}$

$\rm v = 1.26\times10^{26}$ m/sec

Hence the speed of the ion is 1.2×10⁶ m/s.

(B) Radius of cartvature=?

In balancing conditions, Magnetic force is balanced by the centripetal force.

Magnetic force = centripetal force

$\rm{Bqv =\frac{mv^2}{r} }\\\\\\rm r = \frac{mv^2}{bqv} \\\\ \rm r = \frac{1.16\times10^{-26}(1.2\times10^6)^2}{0.0500\times1.6\times10^{-19\times1.2\times10^6}} \\\\\rm r= 2.0\times10^6 m}$

Hence the radius of curvature of a singly charged lithium-ion is 2.0×10⁶ m.

To learn more about the magnetic field refer to the link;

https://brainly.com/question/19542022