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Test: Moving Charges and Magnetism (January 17) - NEET MCQ


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15 Questions MCQ Test Daily Test for NEET Preparation - Test: Moving Charges and Magnetism (January 17)

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Test: Moving Charges and Magnetism (January 17) - Question 1

The force acting on a charge q moving with velocityin a magnetic field is given by

Detailed Solution for Test: Moving Charges and Magnetism (January 17) - Question 1

The magnetic force on a free moving charge is perpendicular to both the velocity of the charge and the magnetic field with direction given by the right hand rule . The force is given by the charge times the vector product of velocity and magnetic field.

Test: Moving Charges and Magnetism (January 17) - Question 2

Find the magnetic force when a charge 3.5C with flux density of 4 units is having a velocity of 2m/s.

Detailed Solution for Test: Moving Charges and Magnetism (January 17) - Question 2

The magnetic force is given by F = q(v x B), where q = 3.5C, v = 2m/s and B = 4 units. Thus we get F = 3.5(2 * 4) = 28 units.

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Test: Moving Charges and Magnetism (January 17) - Question 3

An electron with a speed of 1.8 x 106 m/s is moving in a circular orbit in a uniform magnetic field of 10-4 Wb/m², the radius of the circular path of the electron is​

Detailed Solution for Test: Moving Charges and Magnetism (January 17) - Question 3

4. 0.1063 m

Test: Moving Charges and Magnetism (January 17) - Question 4

A uniform electric field and a uniform magnetic field are produced, pointed in the same direction. An electron is projected with its velocity pointing in the same direction

Detailed Solution for Test: Moving Charges and Magnetism (January 17) - Question 4

As the electron is moving along the direction of the magnetic field, it will experience no magnetic force, but due to an electric force acting on it opposite to the direction of electric field (as it is a negatively charged particle) the velocity of the electron will decreases.

Test: Moving Charges and Magnetism (January 17) - Question 5

A proton enters a magnetic field with a velocity of 1.5 x 105 m/s making an angle of 30° with the field. What is the force experienced by the proton if B = 3.0 T.​

Detailed Solution for Test: Moving Charges and Magnetism (January 17) - Question 5

Test: Moving Charges and Magnetism (January 17) - Question 6

A Charge is fired through a magnetic field. The magnetic force acting on it is maximum when the angle between the direction of motion and magnetic field is

Detailed Solution for Test: Moving Charges and Magnetism (January 17) - Question 6

The force will have a magnitude F=qvB sin q, thus it will be maximum if sin q is maximum. Thus, angle between velocity and magnetic field should be 90o or the charge particle moves perpendicular to the velocity vector.

Test: Moving Charges and Magnetism (January 17) - Question 7

When a charged particle moves in a magnetic field, its kinetic energy always

Detailed Solution for Test: Moving Charges and Magnetism (January 17) - Question 7

When a charged particle enters a magnetic field B its kinetic energy remains constant as the force exerted on the particle is:

is perpendicular to so the work done by =0. This does not cause any change in kinetic energy.

Test: Moving Charges and Magnetism (January 17) - Question 8

A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in the uniform magnetic field. What should be the energy of an alpha particle to describe a circle of the same radius in the same field.

Detailed Solution for Test: Moving Charges and Magnetism (January 17) - Question 8

Kinetic energy of a charged particle,

Radius of the circular path of a charged particle uniform magnetic field is given by

Mass of a proton, mp = m

Mass of an α-particle. mα = 4m

Charge of a proton, qp = e

Charge of an α -particle, qα = 2e

Test: Moving Charges and Magnetism (January 17) - Question 9

An electron is moving in a circular orbit in a magnetic field 2 x 10-3 Wb/m², the time period of electron is

Test: Moving Charges and Magnetism (January 17) - Question 10

The frequency (v) of charged particle, moving at right angles to the magnetic field is independent of

Detailed Solution for Test: Moving Charges and Magnetism (January 17) - Question 10

Frequency (v) = qB/2πm.
Frequency is independent of radius of trajectory of particle and speed of particle.
 

Test: Moving Charges and Magnetism (January 17) - Question 11

The magnetic susceptibility of a material of a rod is 449. Permeability of vacuum is 4π x 10-7 henry/metre. The absolute permeability of the material of the rod (in henry/metre) is

Detailed Solution for Test: Moving Charges and Magnetism (January 17) - Question 11

Test: Moving Charges and Magnetism (January 17) - Question 12

A bar magnet of magnetic moment 2.0 JT-1 lies aligned with the direction of a uniform magnetic field of 0.25 T. What is the work done to turn the magnet so as to align its magnetic moment opposite to the field direction?

Detailed Solution for Test: Moving Charges and Magnetism (January 17) - Question 12

Potential energy of the dipole = 

= 2  2.0  0.25 = 1.0 J

Test: Moving Charges and Magnetism (January 17) - Question 13

A magnet of magnetic moment M and length 2l is bent at its mid-point such that the angle of bending is 60°. Now, the magnetic moment is

Detailed Solution for Test: Moving Charges and Magnetism (January 17) - Question 13

In new sitituation we have

As the length of the magnet is halved, Magnetic Moment 

Resultant Magnetic Moment

Test: Moving Charges and Magnetism (January 17) - Question 14

Two identical thin bar magnets are placed mutually at right angles, such that the north pole of one touches the south pole of the other. The length and the pole strength of each bar are L and m, respectively. The resultant magnetic moment of the system is

Detailed Solution for Test: Moving Charges and Magnetism (January 17) - Question 14

Let the pole strength of the system magnets be m and length N2S1
But

Magnetic moment of the system is 
M - pole strength x N2S1

Test: Moving Charges and Magnetism (January 17) - Question 15

A bar magnet used in a vibration magnetometer is heated, so as to reduce its magnetic moment by 36%. The time period of the magnet (neglecting the changes in the dimension of the magnet) :

Detailed Solution for Test: Moving Charges and Magnetism (January 17) - Question 15

Time period of the  oscillation of the vibration magnetometer is given by

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