Magnetic Effects of Current - NEET Physics Questions
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Magnetic Effects of Current

Question 161: easy

Assertion (A): Work done by magnetic force on any moving charge is zero.


Reason (R): Magnetic force is perpendicular to velocity.


 

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

The magnetic force \( \vec{F}_m = q(\vec{v} \times \vec{B}) \) is always perpendicular to the velocity \( \vec{v} \). Work done by a force is \( W = \vec{F} \cdot \vec{d} \). Since \( \vec{F}_m \perp \vec{v} \), the work done by magnetic force is zero. Thus, both A and R are true, and R correctly explains A.

Question 162: 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.

Question 163: easy

Assertion (A): If a proton and an \( \alpha \)-particle enter a uniform magnetic field perpendicularly, with the same speed, then the time period of revolution of the \( \alpha \)-particle is double than that of proton.


Reason (R): In a magnetic field, the time period of revolution of a charged particle is directly proportional to mass.


 

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

The time period is \( T = \frac{2\pi m}{qB} \). For a proton, \( T_p = \frac{2\pi m_p}{eB} \). For an \( \alpha \)-particle, \( T_\alpha = \frac{2\pi (4m_p)}{2eB} = 2 \frac{2\pi m_p}{eB} = 2T_p \). So, A is true. Reason R (\( T \propto m \)) is true, but it's not the complete explanation for A, as \( T \) also depends on \( q \).

Question 164: easy

Assertion (A): The direction of magnetic moment and orbital angular momentum are opposite to each other for electron.


Reason (R): Electron is negatively charged.


 

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

For an electron, the orbital angular momentum \( vec{L} \) is proportional to \( vec{r} times vec{v} \). The magnetic moment \( vec{mu} \) is \( -frac{e}{2m_e} vec{L} \). The negative sign arises because the electron is negatively charged, causing the direction of the equivalent current to be opposite to the electron's orbital motion. Thus, both A and R are true, and R correctly explains A.

Question 165: easy

Assertion (A): A charged particle moves perpendicular to magnetic field. Its kinetic energy will remain constant but momentum changes.


Reason (R): Magnetic force acts perpendicular to velocity of particle.


 

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

The magnetic force is always perpendicular to the velocity (\( \vec{F} \perp \vec{v} \)). Thus, the work done by the magnetic force is zero (\( W = \vec{F} \cdot \vec{v} t = 0 \)), implying no change in kinetic energy. However, since there is a force, it changes the direction of momentum. Hence, both A and R are true, and R correctly explains A.

Question 166: easy

Assertion (A): A charged particle moving in a magnetic field in general, experiences a force but its kinetic energy remains constant.


Reason (R): Work done by magnetic 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

Magnetic force \( \vec{F} = q(\vec{v} \times \vec{B}) \) is always perpendicular to velocity \( \vec{v} \). Thus, work done \( W = \vec{F} \cdot d\vec{r} = 0 \). By the Work-Energy Theorem, if work done is zero, the kinetic energy remains constant. Both assertion (A) and reason (R) are true, and (R) correctly explains (A).

Question 167: easy

Assertion (A): Electric force between two like charged particles is repulsive but magnetic force between them could be attractive or repulsive or absent depending on the features of their motion.


Reason (R): Magnetic field does not interact with static charges.


 

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

Electric force between like charges is indeed repulsive. Magnetic force between moving charges can be attractive (parallel motion), repulsive (anti-parallel motion), or absent, depending on their relative velocities. Magnetic fields only interact with moving charges, not static ones. Both assertion (A) and reason (R) are true. However, (R) explains why magnetic force requires motion, not the diverse nature (attractive/repulsive) of the force itself.

Question 168: easy

Assertion (A): An electron and a proton enter a uniform magnetic field at right angles to the field with equal velocities, then, deviation of both from the original path will be the same.


Reason (R): In the situation described above, electron and proton will experience magnetic forces of different magnitude.


 

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

The magnitude of magnetic force \( F = qvB sin\theta \) will be the same for an electron and a proton as their charge magnitudes \( q \), velocities \( v \), and magnetic field \( B \) are identical, and \( \theta = 90^\circ \). So (R) is false. The radius of the circular path is \( r = mv/(qB) \). Since \( m_e \ll m_p \), then \( r_e \ll r_p \). Therefore, the electron will deviate more than the proton. So (A) is false. Both (A) and (R) are false.

Question 169: easy

Assertion (A): A charged particle enters a uniform magnetic field with a velocity inclined to the field direction at \( 60^\circ \). The particle will move along a circular path inside the magnetic field.


Reason (R): Magnetic force on a charge inside a magnetic field provides centripetal force for the circular motion of the charge.


 

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

When a charged particle's velocity is inclined to a uniform magnetic field, the component of velocity perpendicular to the field leads to circular motion, while the parallel component leads to linear motion. The combination results in a helical path, which includes a circular component. Thus, (A) is true. The magnetic force \( \vec{F} = q(\vec{v} \times \vec{B}) \) is always perpendicular to \( \vec{v} \), providing the centripetal force \( F_c = qv_{\perp}B \) for the circular motion. So (R) is true and explains the circular aspect of (A).

Question 170: easy

Assertion (A): A current-carrying coil placed in a uniform magnetic field experiences a force which depends on the orientation of plane of the coil relative to the field direction.


Reason (R): A current-carrying conductor placed in a magnetic field experiences a force \( F = I L B sin \theta \).


 

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

A current-carrying coil in a uniform magnetic field experiences a torque \( \tau = NIAB sin \alpha \), where \( \alpha \) is the angle between the area vector and the magnetic field. This torque and the resultant forces clearly depend on the coil's orientation, making (A) true. The force on a segment of a current-carrying conductor is given by \( F = ILB sin \theta \). This fundamental principle explains the origin of the forces acting on the sides of the coil, leading to the overall torque and forces in (A). Both (A) and (R) are true, and (R) is the correct explanation of (A).