Respuesta :
Answer:
[tex]\dfrac{q}{m}=1.32\times 10^{11}\ C/kg[/tex]
Explanation:
It is given that,
Magnetic field, B = 0.00348 T
Radius of circular path, r = 1.17 cm = 0.0117 m
If the charged particles in the beam were accelerated through a potential difference of 110 V, V = 110 V
Let q is the charge and m is the mass.
The centripetal force is balanced by the magnetic force as :
[tex]qvB=\dfrac{mv^2}{r}[/tex]
[tex]Bq=\dfrac{mv}{r}[/tex]...........(1)
Also, [tex]\dfrac{mv^2}{r}=qV[/tex]
[tex]v=\sqrt{\dfrac{2qv}{m}}[/tex]
Put the value of v in equation (1) as :
[tex]B^2q^2r^2=2qmV[/tex]
[tex]\dfrac{q}{m}=\dfrac{2V}{B^2r^2}[/tex]
[tex]\dfrac{q}{m}=\dfrac{2\times 110}{(0.00348)^2\times (0.0117)^2}[/tex]
[tex]\dfrac{q}{m}=1.32\times 10^{11}\ C/kg[/tex]
So, the charge to mass ratio of the charged particles in the beam is [tex]1.32\times 10^{11}\ C/kg[/tex]. Hence, this is the required solution.
Answer:
[tex]\frac{q}{m}=1.33\times 10^{11} \frac{C}{kg}[/tex]
Explanation:
In this question we have given,
magnetic field, [tex]B=0.00348 T[/tex]
radius of circular path, [tex]r=1.17 cm=.0117 m[/tex]
Potential difference, V=110V
let the particle of mass m be moving with velocity, v
We know that force acting on charge particle(q) moving with velocity v in a magnetic field B is given as
[tex]F=qvB[/tex]...........(1)
Similarly centripetal force on charge particle is given as,
[tex]F=\frac{mv^2}{r}[/tex]...............(2)
on comparing equation (1) and equation (2)
[tex]qvB = \frac{mv^2}{r}[/tex]
or
[tex]v = \frac{qBr}{m}[/tex]
Here kinetic energy of charge particle is due to potential difference 110V
Therefore,
Kinetic energy = Energy due to potential difference 110V
[tex]\frac{mv^2}{2}= qV[/tex]...............(3)
put value of v ine equation 3
we got,
[tex]\frac{mr^2B^2q^2}{2m^2}=qV[/tex]
therefore,
[tex]\frac{q}{m}= \frac{2V}{r^2B^2}[/tex].............(4)
put value of V, r and B in equation 4
we got,
[tex]\frac{q}{m}= \frac{2\times 110 V}{(.0117 m)^2\times .00348^2}[/tex]
[tex]\frac{q}{m}=1.33\times 10^{11} \frac{C}{kg}[/tex]
