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 \( q \).
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.
Assertion (A): A charged particle is moving in a circle with constant speed in uniform magnetic field. If we increase the speed of particle to twice, its acceleration will become four times.
Reason (R): A charge particle in circular path with constant speed in magnetic field, acceleration is given by centripetal acceleration. If speed is doubled centripetal acceleration will become four times.
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 centripetal acceleration for a charged particle in a magnetic field is \( a = frac{qvB}{m} \). If speed \( v \) is doubled, then acceleration \( a \) will also double, not quadruple. So, Assertion (A) is false. Similarly, in this context, Reason (R) is also false. Thus, both are false.
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.
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).
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.
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.
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).