Advertisements
Advertisements
प्रश्न
A narrow beam of singly charged potassium ions of kinetic energy 32 keV is injected into a region of width 1.00 cm with a magnetic field of strength 0.500 T, as shown in the figure. The ions are collected at a screen 95.5 cm away from the field region. If the beam contains isotopes of atomic weights 39 and 41, find the separation between the points where these isotopes strike the screen. Take the mass of a potassium ion = A (1.6 × 10−27) kg, where A is the mass number.
Advertisements
उत्तर
Kinetic energy of singly-charged potassium ions = 32 keV
Width of the magnetic region = 1.00 cm
Magnetic field's strength, B = 0.500 T
Distance between the screen and the region = 95.5 cm
Atomic weights of the two isotopes are 39 and 41.
Mass of a potassium ion = A (1.6 × 10−27) kg
For a singly-charged potassium ion K-39 :
Mass of K-39 = 39 × 1.6 × 10−27 kg,
Charge, q = 1.6 × 10−19 C
As per the question, the narrow beam of singly-charged potassium ions is injected into a region of magnetic field.
As
K.E = 32 keV
`1/2 mv^2 = 32 xx 10^3 xx 1.6 xx 10^-19`
`1/2`× 39 × (1.6 × 10−27) × v2 = 32 × 103 × 1.6 × 10−19
v = 4.05 × 105
We know that throughout the motion, the horizontal velocity remains constant.
So, the time taken to cross the magnetic field,
t = `d/v = (0.01)/(4.05xx10^5)`
= 24.7 × 10−9 s
Now, the acceleration in the magnetic field region:
F = qvB = ma
`a = (qvB)/(m)`
= `(1.6xx10^-19xx4.05xx10^5)/(39xx1.6xx10^-27)`
= 5192 × 108 m/s2
Velocity in the vertical direction, vy = at
= 5193.53 × 108 × 24.7 × 10−9
= 12824.24 m/s
Time taken to reach the screen
`d/v = (0.955)/(4.05xx10^5)`
Distance moved vertically in this time
= vy × t
= 12824.24 × 2358 × 10−9
= 3023.95×10-5 m
Vertical distance travelled by the particle inside magnetic field can be found out by using equaton of motion
v2 = 2aS
⇒ (12824.24)2 = 2 × 5192 × 108 × S
⇒ 15.83×10−5 = S
Net display from line
= 15.83×10−5+3023.95×10-5
= 3039.787×10−5 m.
For the potassium ion K-41l :
`1/2`× 41 × 1.6 × 10−27 v2= 32 × 103 × 1.6 × 10−9
⇒ v = 3.94×105 m/s
Similarly, acceleration,
a = 4805 × 108 m/s2
t = time taken for exiting the magnetic field
= 25.4 × 10−9 sec.
vy1= at (vertical velocity)
= 4805 × 108 × 25.4 × 10−9
= 12204.7× 10−9 m/s
Time to reach the screen
= 2423× 10−9 s.
Distance moved vertically
= 12204.7 × 2423 × 10−9
= 2957.1× 10−5
Now,
Vertical distance travelled by the particle inside magnetic field can be found out by using equaton of motion
v2 = 2aS
(12204.7)2 = 2 × 4805 × 108 S
⇒ S = 15.49× 10-5 m
Net distance travelled
= 15.49× 10-5 +2957.1× 10−5
= 2972.68× 10-5 m.
Net gap between K-39 and K-41
= 3039.787×10−5− 2972.68× 10-5
= 67 mm.
APPEARS IN
संबंधित प्रश्न
Write the expression for the force `vecF` acting on a particle of mass m and charge q moving with velocity `vecV` in a magnetic field `vecB` , Under what conditions will it move in (i) a circular path and (ii) a helical path?
A neutron, an electron and an alpha particle, moving with equal velocities, enter a uniform magnetic field going into the plane of the paper, as shown. Trace their paths in the field and justify your answer.
A long horizontal wire P carries a current of 50A. It is rigidly fixed. Another wire Q is placed directly above and parallel to P, as shown in Figure 1 below. The weight per unit length of the wire Q is 0.025 Nm-1 and it carries a current of 25A. Find the distance 'r' of the wire Q from the wire P so that the wire Q remains at rest
A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid air by a uniform magnetic field B. What is the magnitude of the magnetic field?
A beam consisting of protons and electrons moving at the same speed goes through a thin region in which there is a magnetic field perpendicular to the beam. The protons and the electrons
If a charged particle at rest experiences no electromagnetic force,
(a) the electric field must be zero
(b) the magnetic field must be zero
(c) the electric field may or may not be zero
(d) the magnetic field may or may not be zero
If a charged particle moves unaccelerated in a region containing electric and magnetic fields
(a) `vecE "must be perpendicular" to vecB`
(b) `vecv "must be perpendicular" to vecE`
(c) must be perpendicular to v_B
Two particles X and Y having equal charge, after being accelerated through the same potential difference enter a region of uniform magnetic field and describe circular paths of radii R1 and R2 respectively. The ratio of the mass of X to that of Y is ______.
A current of 2 A enters at the corner d of a square frame abcd of side 20 cm and leaves at the opposite corner b. A magnetic field B = 0.1 T exists in the space in a direction perpendicular to the plane of the frame, as shown in the figure. Find the magnitude and direction of the magnetic forces on the four sides of the frame.
Using the formula \[\vec{F} = q \vec{v} \times \vec{B} \text{ and } B = \frac{\mu_0 i}{2\pi r}\]show that the SI units of the magnetic field B and the permeability constant µ0 may be written as N mA−1 and NA−2 respectively.
An electron of kinetic energy 100 eV circulates in a path of radius 10 cm in a magnetic field. Find the magnetic field and the number of revolutions per second made by the electron.
Protons with kinetic energy K emerge from an accelerator as a narrow beam. The beam is bent by a perpendicular magnetic field, so that it just misses a plane target kept at a distance l in front of the accelerator. Find the magnetic field.
A circular coil of radius 2.0 cm has 500 turns and carries a current of 1.0 A. Its axis makes an angle of 30° with the uniform magnetic field of magnitude 0.40 T that exists in the space. Find the torque acting on the coil.
A square coil of edge l and with n turns carries a current i. It is kept on a smooth horizontal plate. A uniform magnetic field B exists parallel to an edge. The total mass of the coil is M. What should be the minimum value of B for which the coil will start tipping over?
A particle of mass m and charge q is projected into a region that has a perpendicular magnetic field B. Find the angle of deviation (figure) of the particle as it comes out of the magnetic field if the width d of the region is very slightly smaller than
(a) `(mv)/(qB)` (b)`(mv)/(2qB)` (c)`(2mv)/(qB)`
A narrow beam of singly-charged carbon ions, moving at a constant velocity of 6.0 × 104m s−1, is sent perpendicularly in a rectangular region of uniform magnetic field B = 0.5 T (figure). It is found that two beams emerge from the field in the backward direction, the separations from the incident beam being 3.0 cm and 3.5 cm. Identify the isotopes present in the ion beam. Take the mass of an ion = A(1.6 × 10−27) kg, where A is the mass number.
A proton is projected with a velocity of 3 × 106 m s−1 perpendicular to a uniform magnetic field of 0.6 T. Find the acceleration of the proton.
A uniform magnetic field of magnitude 0.20 T exists in space from east to west. With what speed should a particle of mass 0.010 g and with charge 1.0 × 10−5 C be projected from south to north so that it moves with uniform velocity?
A particle with a charge of 5.0 µC and a mass of 5.0 × 10−12 kg is projected with a speed of 1.0 km s−1 in a magnetic field of magnitude 5.0 mT. The angle between the magnetic field and the velocity is sin−1 (0.90). Show that the path of the particle will be a helix. Find the diameter of the helix and its pitch.
The velocity of a body of mass 2 kg as a function of time t is given by v(t) = 2t`hat"i" + "t"^2hat"j"`. The force acting on it, at time t = 2 s is given by ______.
