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Test: Moving Charges and Magnetism - 2 - CUET Humanities MCQ


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15 Questions MCQ Test Agriculture Practice Tests: CUET Preparation - Test: Moving Charges and Magnetism - 2

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Test: Moving Charges and Magnetism - 2 - Question 1

Which of the following is not a fact about magnetism?

Detailed Solution for Test: Moving Charges and Magnetism - 2 - Question 1

Repulsion is a sure test of magnetism. Repulsion takes place only between two likes poles of a magnet, whereas attraction takes place between two unlike poles of a magnet and also between a magnet and a magnetic material. So, if a material attracts other material, we cannot be sure whether it is a magnet or not.

Test: Moving Charges and Magnetism - 2 - Question 2

An electron beam is moving between two parallel plates having an electric field 1.125 x 10-6 N/m. A magnetic field 3 x 10-10 T is also applied so that the beam does not deflect. The velocity of the electron is

Detailed Solution for Test: Moving Charges and Magnetism - 2 - Question 2

As the electron moves undeflected between the electric field and the magnetic field, so the resultant force on the electron is zero.
qE = qvB
v = E/B

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Test: Moving Charges and Magnetism - 2 - Question 3

An α-particle with a specific charge of 2.5 × 107 C-kg-1 moves with a speed of 2 × 105 ms-1 in a perpendicular magnetic field of 0.05 T. Then, the radius of the circular path described by it is

Detailed Solution for Test: Moving Charges and Magnetism - 2 - Question 3

r = mv/Bq
⇒ 
 cm = 16 cm

Test: Moving Charges and Magnetism - 2 - Question 4

A direct current is sent through a helical wire coil or spring. The coil

Detailed Solution for Test: Moving Charges and Magnetism - 2 - Question 4

If a direct current is sent through a helical wire coil, the sense of the current in the adjacent turns will be the same. We know that parallel currents attract each other, so the coil will tend to get shorter.

Test: Moving Charges and Magnetism - 2 - Question 5

The direction of the force experienced by a charged particle, moving with a velocity v in a uniform magnetic field B, is

Detailed Solution for Test: Moving Charges and Magnetism - 2 - Question 5

The force on the moving charge particle due to the magnetic field is:

So, the direction of the force experienced by a charged particle, moving with a velocity v in a uniform magnetic field B, is perpendicular to both the velocity of the charged particle and the direction of the magnetic field.

Test: Moving Charges and Magnetism - 2 - Question 6

When an electron moves in a magnetic field in a direction perpendicular to the field, then its path will be

Detailed Solution for Test: Moving Charges and Magnetism - 2 - Question 6

When an electron moves in a magnetic field in a direction perpendicular to the field, then its path will be circular as the force exerted by the magnetic field is perpendicular to both the magnetic field vector and the velocity vector. The necessary centripetal force is provided by the force due to the magnetic field.

Test: Moving Charges and Magnetism - 2 - Question 7

An electron of mass m and charge q is travelling with a speed v along a circular path of radius r at a right angle to a uniform magnetic field of intensity B. If the speed of the electron is doubled and the intensity of magnetic field is halved, the resulting path would have a radius

Detailed Solution for Test: Moving Charges and Magnetism - 2 - Question 7

Radius of path of electron r = mv/Bq 
m and q remain unchanged. (as r1 = r)
So, 

Test: Moving Charges and Magnetism - 2 - Question 8

A straight section PQ of a circuit lies along the x-axis from x = - a/2 to x = a/2 and carries a current I. The magnetic field due to the section PQ at point x = +a will be

Detailed Solution for Test: Moving Charges and Magnetism - 2 - Question 8

The point x = +a lies along the line of the straight section PQ of the circuit. Hence, the magnetic field at point x = a is zero.

Test: Moving Charges and Magnetism - 2 - Question 9

A power line lies along the East-West direction and carries a current of 10 A. The force per unit length due to the Earth's magnetic field of 10–4 T is

Detailed Solution for Test: Moving Charges and Magnetism - 2 - Question 9

Force F on a wire of length l = iLB
Therefore, force per unit length of the wire = F/L = iB = 10–4 × 10 = 10–3 Nm–1

Test: Moving Charges and Magnetism - 2 - Question 10

A circular coil of radius r, having n number of turns and carrying a current I, produces a magnetic field of magnitude B at its centre. B can be doubled by

Detailed Solution for Test: Moving Charges and Magnetism - 2 - Question 10

The magnetic field at the centre of the coil is given by:

Magnetic field can be doubled by changing n to 2n and by keeping the current same.
Hence, the correct choice is (4).

Test: Moving Charges and Magnetism - 2 - Question 11

A charged particle is released from rest in a region of steady and uniform electric and magnetic fields, which are parallel to each other. The particle will move in a

Detailed Solution for Test: Moving Charges and Magnetism - 2 - Question 11

If a charged particle is released from rest in a region of steady and uniform electric and magnetic fields, which are parallel to each other, then there is no force exerted by the magnetic field on the charged particle as initially, the charged particle is at rest and after that, the direction of the velocity of the charged particle is along the direction of the charged particle i.e. , where the force on the charged particle will be exerted by the electric field i.e. .
Direction of the force exerted by the electric field will be the same as that of electric field, if the charged particle is positive.
Direction of the force exerted by the electric field will be opposite to the electric field, if the charged particle is negative.

Test: Moving Charges and Magnetism - 2 - Question 12

An electron of charge e moves in a circular orbit of radius r around the nucleus at a frequency v. The magnetic moment associated with the orbital motion of the electron is

Detailed Solution for Test: Moving Charges and Magnetism - 2 - Question 12

The charge passing per second through any point of the path is v times the charge of the electron, i.e. i = ev. If A is the area of the orbit, the magnetic moment is:
m = iA = evπr2
Hence, the correct choice is (1).

Test: Moving Charges and Magnetism - 2 - Question 13

Which of the following statements are correct for induced magnetism?
(a) It is defined as the magnetism acquired by a magnetic material kept near a magnet.
(b) A similar polarity develops on the magnet and the material in its vicinity.
(c) It precedes attraction.
(d) It is permanent.

Detailed Solution for Test: Moving Charges and Magnetism - 2 - Question 13

Induced magnetism is defined as the magnetism acquired by a magnetic material kept near a magnet, and it precedes attraction.

Test: Moving Charges and Magnetism - 2 - Question 14

A particle of charge q and mass m moves in a circular orbit of radius r with an angular speed ω. The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on

Detailed Solution for Test: Moving Charges and Magnetism - 2 - Question 14

Magnetic moment M = Current x Area =  

Test: Moving Charges and Magnetism - 2 - Question 15

A wire of length L metres carrying a current I amperes is bent in the form of a circle. The magnitude of the magnetic moment is

Detailed Solution for Test: Moving Charges and Magnetism - 2 - Question 15

Magnetic moment m = AI = πr2I, where r is the radius of the circular loop.
Now, the circumference of the circle = length of the wire, i.e.
2πr = L
or r2 = 
Therefore, m = r2I = 
Hence the correct choice is (4)

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