Magnetic Effects Of Current Test

Result

Magnetic Effects Of Current Test - 7

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Question 1
1 / -0

A proton, a deutron and an a-particle are accelerated through same potential difference and then they enter a normal uniform magnetic field. The ratio of their kinetic energies will be-

A
2 : 1 : 3

B
1 : 1 : 2

C
1 : 1 : 1

D
1 : 2 : 4

Solution
Question 2
1 / -0

A proton of energy 8eV is moving in a circular path in a uniform magnetic field. The energy of an a-particle moving in the same magnetic field and along the same path will be-

A
4eV

B
2eV

C
8eV

D
6eV

Solution
Question 3
1 / -0

An electron is revolving in a circular path of radius 2×10^–10 m with a speed of 3×10⁶ m/s. The magnetic field at the centre of circular path will be-

A
1.2 T

B
2.4 T

C
0

D
3.6 T

Solution
Question 4
1 / -0

An a particle travels at an angle of 30º to a magnetic field 0.8 T with a velocity of 10⁵ m/s. The magnitude of force will be-

A
1.28 × 10^–14 N

B
(1.28)√3 × 10^–4 N

C
1.28 × 10^–4 N

D
(12.8)√3 × 10^–4 N

Solution

F = qvB sinθ
Question 5
1 / -0

A beam of protons is moving horizontally towards you. As it approaches, it passes through a magnetic field directed downward. The beam deflects-

A
to your left side

B
to your right side

C
does not deflect

D
nothing can be said

Solution

The direction of which is towards the right. Thus the beam deflects to your right side
Question 6
1 / -0

If a particle moves in a circular path in clockwise direction after entering into a downward vertical magnetic field. The charge on the particle is-

A
positive

B
negative

C
nothing can be said

D
neutral

Solution

The particle is moving clockwise which shows that force on the particle is opposite to given by right hand palm rule of fleming left hand rule. These two laws are used for positive charge. Here since laws are disobeyed, we can say that charge is negative.
Question 7
1 / -0

In the example above, after how much time, particle comes to the starting point for the first time. (mass of particle = m)

A

B

C
Never

D
It will leave the circular path before coming to the starting point

Solution

The point lies at the circumference hence it will come back after a time period T
Question 8
1 / -0

A current of 2.0 amp is flowing through a wire of length 50 cm. If this wire be placed at an angle of 60º with the direction of a uniform magnetic field of 5.0 × 10^–4 N/Am the force on the wire will be

A
4.33 × 10^–4 N

B
2.50 × 10^–4 N

C
5.0 × 10^–4 N

D
2.33 × 10^–4 N

Solution

The magnetic force on a current carrying wire of length L, placed in a magnetic field B at an angleθ with the field is given by

According to the flemings left - hand rule, this force will act perpendicular to both the wire and the magnetic field.
Question 9
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A particle of mass m and charge q moves with a constant velocity v along the positive x direction. It enters a region containing a uniform magnetic field B directed along the negative z direction, extending from x = a to x = b. The minimum value of v required so that the particle can just enter the region x > b is

A
qb B/m

B
q(b−a)B/m

C
qaB/m

D
q(b+a)B/2m

Solution

In the figure, the z-axis points out of the paper, and the magnetic field is directed into the paper, existing in the region between PQ and RS. The particle moves in a circular path of radius r in the magnetic field. It can just enter the region x > b for r≥(b−a)
Question 10
1 / -0

A particle with charge Q, moving with a momentum p, enters a uniform magnetic field normally. The magnetic field has magnitude B and is confined to a region of width d, where d < p/BQ.The particle is deflected by an angle θ in crossing the field.

A
sin θ = BQd/p

B
sin θ = p/BQd

C
sin θ = Bp/Qd

D
sin θ = pd/BQ

Solution