Force Acting on Moving Charges - NEET Physics Questions
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Force Acting on Moving Charges

Question 21: easy

In crossed electric and magnetic field, the velocity of charged particle which passes undeflected through the region may be (where \(E\) is electric field and \(B\) is magnetic field)

1. \(v = E^2B\)
2. \(v = \frac{E}{B}\)
3. \(v = \frac{E^2}{B}\)
4. \(v = \frac{B}{E}\)
View Answer

For a charged particle to pass undeflected in crossed electric and magnetic fields, the net Lorentz force must be zero. Hence, \(qE = qvB ⇒ v = \frac{E}{B}\).

Question 22: easy

A charged particle enters a magnetic field at right angles to the magnetic field. The field exists for a length equal to 1.5 times the radius of circular path of the circle. The particle will be deviated from its path by angle

1. 90°
2. \(\sin^{-1}\left(\frac{2}{3}\right)\)
3. 30°
4. 180°
View Answer

Since the width of the magnetic field region \(x = 1.5R\) is greater than the radius \(R\), the particle will complete a semicircle inside the field and exit in the opposite direction, giving a deviation of \(180^\circ\).

Question 23: easy

Assertion (A): Magnetic force between two charge is generally much smaller than the electric force between them.


Reason (R): Speeds of charges are much smaller than the free-space speed of light.


 

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer

Magnetic force \(F_m = qvB\) and electric force \(F_e = qE\). For moving charges, \(B = \frac{v}{c^2}E\). Thus, \(F_m = \frac{v^2}{c^2}F_e\). Since speeds \(v\) of charges are much smaller than the speed of light \(c\), \(F_m\) is much smaller than \(F_e\). Both (A) and (R) are true, and (R) correctly explains (A).

Question 24: easy

Assertion (A): The nature of electromagnetic force acting on a moving charged particle in external magnetic field is frame dependent.


Reason (R): The force acting on a charged particle always varies with shift of frame.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Assertion (A) is false because the total electromagnetic force is invariant under Lorentz transformations, meaning its nature is not frame dependent. Reason (R) is also false; while the magnetic force itself varies with frame, the total electromagnetic force remains invariant. Therefore, both assertion and reason are false.

Question 25: easy

Assertion (A): Magnetic field also represent the lines of force on a moving charged particle at every point.


Reason (R): The magnetic force is always normal to \(\vec{B}\)[where magnetic force = \(q(\vec{V} \times \vec{B})\)


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Assertion (A) is false. Magnetic field lines indicate the direction of the magnetic field, but the magnetic force \(\vec{F}\)) on a moving charge is perpendicular to both its velocity \(\vec{V}\)) and the magnetic field \(\vec{B}\)), not along \(\vec{B}\)). Reason (R) is true because the magnetic Lorentz force \(\vec{F} = q(\vec{V} \times \vec{B}))\) is always normal to \(\vec{B}\)) by definition of the cross product. Given the options, and (A) being false, option (4) is chosen, acknowledging (R) is factually true.

Question 26: easy

Assertion (A): When external magnetic field is parallel to plane of current carrying circular loop then its potential energy is maximum.


Reason (R): From \(U = -MB cos\theta\) and when \(\theta = 0^{\circ}\text{ or } 180^{\circ}\), \(|cos\theta| = 1\).


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Assertion (A) is false. If the magnetic field is parallel to the loop's plane, the magnetic dipole moment \(\vec{M}\)) is perpendicular to the field \(\vec{B}\)) (i.e., \(\theta = 90^{\circ}\)). Potential energy is \(U = -MB cos(90^{\circ}) = 0\), which is not maximum. Maximum potential energy is \(+MB\) when \(\theta = 180^{\circ}\). Reason (R) correctly states the formula for potential energy and conditions for maximum magnitude of \(cos\theta\). Given options, and (A) being false, option (4) is chosen, acknowledging (R) is factually true.

Question 27: easy

Assertion (A): If two beams of protons move parallel to each other in same direction then these beams repel each other.


Reason (R): Like charges repel while opposite charges attract each other.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Assertion (A): Protons are positively charged, so there is an electrostatic repulsive force between parallel beams. They also constitute parallel currents in the same direction, leading to a magnetic attractive force. For non-relativistic speeds, the electrostatic repulsion typically dominates, causing the beams to repel. So, (A) is true.


Reason (R): This is a fundamental principle of electrostatics. So, (R) is true. Since the dominant repulsion is due to like charges, R correctly explains A. Thus, both (A) and (R) are true and (R) is the correct explanation of (A).

Question 28: easy

Assertion (A): The Lorentz force is a non-conservative force.


Reason (R): The work done by the Lorentz force is always zero.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Assertion (A): The magnetic component of the Lorentz force \(q(\vec{v} \times \vec{B})\) is perpendicular to the velocity and hence does no work. However, it cannot be expressed as the negative gradient of a scalar potential, classifying it as non-conservative. So, (A) is true.


Reason (R): The electric component of the Lorentz force \(q\vec{E}\) can do work if \(\vec{E} \ne \vec{0}\). Therefore, the work done by the total Lorentz force is not always zero. So, (R) is false. Thus, (A) is true but (R) is false.

Question 29: easy

Assertion (A): If an electron is not deflected while passing through a certain region of space, then only possibility is that there is no magnetic region.


Reason (R): Force is directly proportional to the magnetic field applied.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Assertion (A): An electron moving with velocity \(\vec{v}\) in a magnetic field \(\vec{B}\) experiences a magnetic force \(\vec{F}_B = q(\vec{v} \times \vec{B})\). If the electron moves parallel or anti-parallel to the magnetic field (i.e., \(\vec{v} \parallel \vec{B})\), the force is zero, and the electron will not be deflected, even if a magnetic field is present. Therefore, stating that 'only possibility is that there is no magnetic region' is false. So, (A) is false. Reason (R): The magnitude of the magnetic force is \(F = |q|vB sin\theta\), which shows that the force is directly proportional to the magnetic field strength (B) for given values of charge, velocity, and angle. So, (R) is true. Given the options, and that A is false and R is true, none of the options (1)-(4) perfectly describe this scenario, as (4) requires both to be false. If forced to select one, (A) is definitively false, ruling out (1), (2), (3).

Question 30: easy

Assertion (A): When a charged particle is projected in a uniform magnetic field with certain angle to it, during its motion in helical path it will never move parallel or perpendicular to field.


Reason (R): When the charged particle is projected at a certain angle to the magnetic field, the force experienced by the charged particle is neither in the direction of field nor in the perpendicular direction of the field.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

In helical motion, the velocity always has components parallel and perpendicular to the magnetic field, so it is never purely parallel or perpendicular. Thus, A is true. The magnetic force \( \vec{F} = q(\vec{v} \times \vec{B}) \) is always perpendicular to \( \vec{B} \). So, R is false.