Advertisements
Advertisements
Question
A magnetic field set up using Helmholtz coils is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is 9.0 × 10–5 V m–1, make a simple guess as to what the beam contains. Why is the answer not unique?
Solution
Magnetic field, B = 0.75 T
Accelerating voltage, V = 15 kV = 15 × 103 V
Electrostatic field, E = 9 × 10–5 V m−1
Mass of the electron = m
Charge of the electron = e
Velocity of the electron = v
Kinetic energy of the electron = eV
`1/2"mv"^2` = eV
∴ `"e"/"m" = "v"^2/(2"V")` ..........(1)
Since the particle remains undeflected by electric and magnetic fields, we can infer that the force on the charged particle due to electric field is balancing the force on the charged particle due to magnetic field.
∴ eE = evB
v = `"E"/"B"` ............(2)
Putting equation (2) in equation (1), we get
`"e"/"m" = 1/2(("E"/"B")^2)/"V" = "E"^2/(2"VB"^2)`
= `(9.0 xx 10^-5)^2/(2 xx 15000 xx (0.75)^2)`
= 4.8 × 107 C/kg
This value of specific charge e/m is equal to the value of deuteron or deuterium ions. This is not a unique answer. Other possible answers are He++, Li++, etc.
APPEARS IN
RELATED QUESTIONS
A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.
Define one tesla using the expression for the magnetic force acting on a particle of charge q moving with velocity \[\vec{v}\] in a magnetic field \[\vec{B}\] .
A proton enters into a magnetic field of induction 1.732 T, with a velocity of 107 m/s at an angle 60° to the field. The force acting on the proton is e = 1.6 × 10-19 C, sin 60° = cos 30° = `sqrt3/2`
For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by,
B = `(μ_0"IR"^2"N")/(2("x"^2 + "R"^2)^(3/2))`
(a) Show that this reduces to the familiar result for field at the centre of the coil.
(b) Consider two parallel co-axial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by, B = `0.72 (μ_0"NI")/"R"` approximately.
[Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]
A solenoid 60 cm long and of radius 4.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g lies inside the solenoid (near its centre) normal to its axis; both the wire and the axis of the solenoid are in the horizontal plane. The wire is connected through two leads parallel to the axis of the solenoid to an external battery which supplies a current of 6.0 A in the wire. What value of current (with appropriate sense of circulation) in the windings of the solenoid can support the weight of the wire? (g = 9.8 m s–2)
A magnetic field exerts no force on
A charged particle would continue to move with a constant velocity in a region wherein ______.
- E = 0, B ≠ 0.
- E ≠ 0, B ≠ 0.
- E ≠ 0, B = 0.
- E = 0, B = 0.
At a certain place the angle of dip is 30° and the horizontal component of earth’s magnetic field is 0.5 G. The earth’s total magnetic field (in G), at that certain place, is ______.
Distinguish between the forces experienced by a moving charge in a uniform electric field and in a uniform magnetic field. (Any two points)
