NCERT Based Test: Motion in Combined Electric & Magnetic Fields
15 Questions MCQ Test Physics Class 12 | NCERT Based Test: Motion in Combined Electric & Magnetic Fields
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A charged particle with charge q enters a region of constant, uniform and mutually orthogonal fields 
with a velocity 
perpendicular to both 
, and comes out without any change in its magnitude or direction. Then
with a velocity perpendicular to both , and comes out without any change in its magnitude or direction. Then When 
are perpendicular and velocity has no change then qE = qvB i.e. v= E/B. The two forces oppose each other so, v is along 
i.e., 
.
As 
are perpendicular to each other 
A charged particle would continue to move with a constant velocity in a region wherein, which of the following conditions is not correct?
The cyclotron frequency vc is given by
In cyclotron the centripetal force is balanced by magnetic force, then
qvB = mn2/r ⇒ qBr/m = v
where v = 2πrvc ∴ vc = qB/2πm
A charged particle is moving in a cyclotron, what effect on the radius of path of this charged particle will occur when the frequency of the radio frequency field is doubled?
As frequency of revolution in a cyclotron
vC = Bq/2πm is independent of r.
So the radius of path in the dees will remain unchanged. When the frequency is changed.
A cubical region of space is filled with some uniform electric and magnetic field. An electron enters the cube across one of its faces with velocity v and a positron enters via opposite face with velocity - v. At this instant, which one of the following is not correct?
A cyclotron is operated at an oscillator frequency of 12 MHz and has a dee radius R = 50 cm. What is the magnitude of the magnetic field needed for a proton to be accelerated in the cyclotron?
If an electron is moving in a magnetic field of 5.4 x 10-4 T on a circular path of radius 32 cm having a frequency of 2.5 MHz, then its speed will be
Here, B = 5.4 x 10-4 T;
r = 32 cm = 32 x 10-2 m; v = 2.5 MHz = 2.5 x 106Hz
The speed of electron on circular path
v = r x 2πv = 32 x 10-2 x 2 x 3.14 x 2.5 x 106
= 502.4 x 104 = 5.024 x 106 m s-1
A proton and an α-partide enter in a unifoi magnetic field perpendicularly with same speed. The ratio of time periods of both particle (TP/Tα) will be
The time period of revolution of a charged particle in a magnetic field is
T = 2πm/Bq
For proton, mP = m, qP = q ; ∴ TP = 2πm/Bq
Now, for α-particle, mα = 4m, qα = 2q
∴ Tα = (2π(4m))/(B(Bq)) = 2(2πm/Bq) ⇒ TP/Tα = 1/2
A proton, a deutron and an α-particle with same, kinetic energy enter perpendicularly in a uniform magnetic field, then the ratio of radii of their circular paths is
∵ r =
∴ r ∝ √m/q
Hence, rp : rd ; rα = 
= 
If an electron enters with a velocity 
into a cubical region in a uniform electromagnetic field and the orbit of the electron is found to spiral down inside the cube in plane parallel to x-y plane then the configuration of electric field and magnetic field is
The orbit of the electron spirals down inside the cube in the plane parallel to x−y plane. Therefore 
must be along +z direction and 
should be along x−axis.
The energy of emergent protons in MeV from a cyclotron having radius of its dees 1.8 m and applied magnetic field 0.7T is (mass of proton = 1.67×10−27 kg):
= 1.22 x 10-11 j = 76 MeV
Which of the following is not correct about cyclotron?
The charged particles and ions in cyclotron can move along circular path only.
The operating magnetic field for accelerating protons in a cyclotron oscillator having frequency of 12 MHz is (q = 1.6 x 10-19C, mp = 1.67 x 10-27 kg and 1 MeV = 1.6 x 10-13 j)
In the question number 34, the kinetic energy (in MeV) of the proton beam produced by the accelerator is (radius of dees = 60 cm)
Final velocity of proton is
v = r x 2πv = 0.6 x 2 x 3.14 x 12 x 106 = 4.5 x 107 m s-1
= 10.6 MeV
An electron is moving in a cyclotron at a speed of 3.2 x 107 m s-1 in a magnetic field of 5 x 10-4 T perpendicular to it. What is the frequency of this electron?
(q = 1.6 x 10-19 C, me = 9.1 x 10-31 kg)
v = 3.2 x 107 m s-1 ;B = 5 x 10-4 T
The frequency of electron is
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