Moving Charges and Magnetism Test

Result

Moving Charges and Magnetism Test - 35

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Weekly Quiz Competition

Question 1
4 / -1

A
Always zero

B
Never zero

C

D

Solution
Question 2
4 / -1

A proton of mass m and charge +e is moving in a circular orbit in a magnetic field with energy 1 MeV. What should be the energy of α–particle (mass = 4m and charge = +2e), so that it can revolve in the path of same radius

A
1 MeV

B
4 MeV

C
2 MeV

D
0.5 MeV

Solution
Question 3
4 / -1

A
No change

B
Reduces to r/2

C
Increases to 2r

D
Stops moving

Solution
Question 4
4 / -1

A proton and an α–particle enter a uniform magnetic field perpendicularly with the same speed. If proton takes 25 μ sec to make 5 revolutions, then the periodic time for the α–particle would be

A
50 μ sec

B
25 μ sec

C
10 μ sec

D
5 μ sec

Solution
Question 5
4 / -1

A proton (mass = 1.67 × 10^–27kg and charge = 1.6 × 10^–19 C) enters perpendicular to a magnetic field of intensity 2 weber/m² with a velocity 3.4 × 10⁷m/sec. The acceleration of the proton should be

A
6.5 × 10¹⁵ m/sec²

B
6.5 × 10¹³ m/sec²

C
6.5 × 10¹¹ m/sec²

D
6.5 × 10⁹ m/sec²

Solution
Question 6
4 / -1

A charged particle is moving in a circular orbit of radius 6 cm with a uniform speed of 3 × 10⁶ m/s under the action of a uniform magnetic field 2 × 10^–4 wb/m² at right angles to the plane of the orbit. The charge to mass ratio of the particle is

A
5 × 10⁹ C/kg

B
2.5 × 10¹¹ C/kg

C
5 × 10¹¹ C/kg

D
5 × 10¹² C/kg

Solution
Question 7
4 / -1

A uniform magnetic field acts at right angles to the direction of motion of electrons. As a result, the electron moves in a circular path of radius 2 cm. If the speed of the electrons is doubled, then the radius of the circular path will be

A
2.0 cm

B
0.5 cm

C
4.0 cm

D
1.0 cm

Solution
Question 8
4 / -1

A
25 keV

B
50 keV

C
200 keV

D
100 keV

Solution
Question 9
4 / -1

A current I flowing through the loop as shown in figure. The magnetic field at centre O is

A

B

C

D

Solution
Question 10
4 / -1

A

B

C

D

Solution