A proton is accelerated through a potential difference V, subjected to a
uniform magnetic field acting normal to the velocity of the proton. If the
potential difference is doubled, how will the radius of the circular path
described by the proton in the magnetic field change ?
155/2/17
P.T
Answers
Answer:
The radius of the circle is given by :
R=\dfrac{mv}{qB}R=
qB
mv
...(i)
Where m,v and q are mass, velocity of proton and charge of proton,
The proton is accelerated with potential difference of V, it's KE is given as K.E.= qVK.E.=qV
\dfrac{1}{2}mv^2=qV \Rightarrow v= \sqrt{\dfrac{2qV}{m}}
2
1
mv
2
=qV⇒v=
m
2qV
...(ii)
substituting value of vv in equation (i) from equation (ii)
R=\dfrac{m\sqrt{\dfrac{2qV}{m}}}{qB}=\dfrac{\sqrt{\dfrac{2mV}{q}}}{B}R=
qB
m
m
2qV
=
B
q
2mV
...(iii)
from geometry Rsin\alpha=d\Rightarrow sin\alpha=\dfrac{d}{R}Rsinα=d⇒sinα=
R
d
Substituting value of R from equation (iii)
sin\alpha=dB\sqrt{\dfrac{q}{2mV}}sinα=dB
2mV
q
.
Explanation:
R= MV/QB
mv= √2mqV, where V is potential difference
R∝ √V
Radius will become √2 times.