An electron is moving in air at right angles to uniform magnetic field. The diagram below shows the path of the electron. The electron is slowing down.
Which one of the following correctly gives the direction of motion of the electron and the direction of the magnetic field ?
Direction of motion
Direction of magnetic field
1.
clockwise
into plane of paper
2.
out of plane of paper
3.
anti-clockwise
4.
Rank the value of ∮B→·dl→ for the closed paths shown in figure from the smallest to largest:
a, b, c, d
a, c, d, b
a, d, c, b
a, c, b, d
Electrons moving with different speeds enter in uniform magnetic field in a direction perpendicular to the field. They will move along circular paths:
of same radius
with larger radius for the faster electrons
with smaller radius for the faster electrons
either 2 or 3 depending on the magnitude of the magnetic field
Electrons moving with different speeds enter in uniform magnetic field in a direction perpendicular to the field. If they move along circular paths, then the time periods of rotation will be:
same for all electrons
greater for the faster electrons
smaller for the faster electrons
Consider the three closed loops drawn using solid line in the magnetic field (magnetic field lines are drawn using dotted line) of an infinite curent-carrying wire normal to the plane of paper as shown.
Rank the line integral of the magnetic field along each path in order of increasing magnitude:
1 > 2 > 3
3. 1 = 2 = 3
2. 1 = 3 > 2
3 > 2 > 1
Only the current inside the Amperian loop contributes in:
finding magnetic field at any point on the Ampere's loop
line integral of magnetic field
in both of the above
in neither of them
An electron and a proton are moving under the influence of mutual forces. In calculating the change in the kinetic energy of the system during motion, one ignores the magnetic force of one on another. This is because:
the two magnetic forces are equal and opposite, so they produce no net effect.
the magnetic forces do no work on each particle.
the magnetic forces do equal and opposite (but nonzero) work on each particle.
the magenetic forces are necessarily negligible.
Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field B = B0k^.
They have equal z-components of momenta.
They must have equal charges.
They necessarily represent a particle-antiparticle pair.
The charge to mass ratio satisfy: (em)1+(em)2=0
Biot-Savart law indicates that the moving electrons (velocity v) produce a magnetic field B such that:
B ⊥ v.
B || v.
it obeys inverse cube law.
it is along the line joining the electron and point of observation.
The magnetic moment does not change.
The magnitude of B at (0, 0, z), z >>R increases.
The magnitude of B at (0, 0, z), z >>R is unchanged.
The electron path will be circular about the axis.
The electron will experience a force at 45° to the axis and hence execute a helical path.
The electron will continue to move with uniform velocity along the axis of the solenoid.
In a cyclotron, a charged particle:
undergoes acceleration all the time.
speeds up between the dees because of the magnetic field.
speeds up in a dee.
slows down within a dee and speeds up between dees.
A circular current loop of magnetic moment M is in an arbitraryorientation in an external magnetic field B. The work is done to rotatethe loop by 30° about an axis perpendicular to its plane is:
MB
3MB2
MB/2
zero
negative.
positive.
increases with the quantum number n.
Consider a wire carrying a steady current, I placed in a uniform magnetic field B perpendicular to its length. Consider the charges inside the wire. It is known that magnetic forces do no work. This implies that,
(a) the motion of charges inside the conductor is unaffected by B since they do not absorb energy.(b) some charges inside the wire move to the surface as a result of B.(c) if the wire moves under the influence of B, no work is done by the force.(d) if the wire moves under the influence of B, no work is done by the magnetic force on the ions assumed fixed within the wire.
(b, c)
(a, d)
(b, d)
(c, d)
Two identical current-carrying coaxial loops, carry current I in an opposite sense. A simple amperian loop passes through both of them once. Calling the loop as C,a. ∮B.dl=∓2μ0I.b. the value of ∮B.dl is independent of the sense of C.c. there may be a point on C where B and dl are perpendicular.d. B vanishes everywhere on C.
A cubical region of space is filled with some uniform electric and magnetic fields. An electron enters the cube across one of its faces with velocity v and a positron enters via the opposite face with velocity - v. At this instant,
(a) the electric forces on both the particles cause identical accelerations.(b) the magnetic forces on both the particles cause equal accelerations.(c) both particles gain or lose energy at the same rate.(d) the motion of the centre of mass (CM) is determined by B alone.
(a, b, c)
(a, c, d)
(b, c, d)
A charged particle would continue to move with a constant velocity in a region wherein,
(a) E = 0, B ≠ 0.(b) E ≠ 0, B ≠ 0.(c) E ≠ 0, B = 0.(d) E = 0, B = 0.
(a, c)
A wire of length L meter carrying a current of I ampere is bent in the form of a circle. Its magnetic moment is,
IL24 A-m2
I×πL24 A-m2
2IL2π A-m2
IL24π A-m2
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 horizontal magnetic field B (shown in the figure). What is the magnitude of the magnetic field?
0.65 T
0.77 T
0.44 T
0.20 T
The radius of the path of an electron and frequency (mass 9×10-31 kg and charge 1.6×10-19 C) moving at a speed of 3 ×107 m/s in a magnetic field of 6 × 10–4 T perpendicular to it are respectively:
24 cm, 4 MHz
22 cm, 4 MHz
28 cm, 2 MHz
26 cm, 2 MHz
An electron mass 9×10-31 kg and charge 1.6×10-19 C is moving at a speed of 3 ×107 m/s in a magnetic field perpendicular to it. The energy of the electron in keV is: ( 1 eV = 1.6 × 10–19 J)
2.5 keV
3.5 keV
20 keV
3.0 keV
A cyclotron’s oscillator frequency is 10 MHz. The radius of its ‘dees’ is 60 cm, what should be the operating magnetic field for accelerating protons and the kinetic energy (in MeV) of the proton beam produced by the accelerator?
0.66 T, 6 MeV
0.33 T, 6 MeV
0.66 T, 7 MeV
0.33 T, 7 MeV
An element ∆l=∆xi^ is placed at the origin and carries a large current I =10 A (as shown in the figure). What is the magnetic field on the y-axis at a distance of 0.5 m? (∆x = 1 cm).
1. 6×10-8 T2. 4×10-8 T3. 5×10-8 T4. 5.4×10-8 T
A straight wire carrying a current of 12 A is bent into a semi-circular arc of radius 2.0 cm as shown in the figure. Consider the magnetic field B at the centre of the arc. What is the magnetic field due to the straight segments?
1. 02. 1.2×10-4 T3. 2.1×10-4 T4. None of these
Consider a tightly wound 100 turn coil of radius 10 cm, carrying a current of 1 A. What is the magnitude of the magnetic field at the centre of the coil?
1. 8.2×10-4 T2. 4.6×10-4 T3. 5.2×10-4 T4. 6.2×10-4 T
The figure shows a long straight wire of a circular cross-section (radius a) carrying steady current I. The current I is uniformly distributed across this cross-section. If the magnetic field in the region (r < a) is B1 and for (r > a) is B2, then B1B2 is:
1. ar2. a2r23. r2a24. ra
A solenoid of length 0.5 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 5 A. What is the magnitude of the magnetic field inside the solenoid?
1. 6.28×10-3 T2. 3.14×10-3 T3. 2.72×10-3 T4. 5.17×10-3 T
The horizontal component of the earth’s magnetic field at a certain place is 3.0 ×10–5 T and the direction of the field is from the geographic south to the geographic north. A very long straight conductor is carrying a steady current of 1 A. What is the force per unit length on it when it is placed on a horizontal table and the direction of the current east to west?
1. 3×10-5 N m-12. 2×10-7 N m-13. 04. 5×10-7 N m-1
A 100 turn closely wound circular coil of radius 10 cm carries a current of 3.2 A. The magnetic field at the centre of the coil is:
1. 4×10-3 T2. 6×10-3 T3. 2×10-3 T4. 5×10-3 T
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