A charged particle is moving on circular path with velocity v in a uniform magnetic field B, if the velocity of the charged particle is doubled and strength of Magnetic field is halved, then radius becomesa)8 timesb)4 timesc)2 timesd)16 timesCorrect answer is option 'B'. Can you explain this answer?

Question

Answer

Explanation:
When a charged particle moves in a magnetic field, it experiences a magnetic force that acts perpendicular to both the velocity of the particle and the magnetic field. This force causes the particle to move in a circular path.

Given:
Initial velocity (v) of the charged particle
Initial magnetic field strength (B)
Final velocity of the charged particle = 2v
Final magnetic field strength = B/2

Formula:
The force experienced by a charged particle moving in a magnetic field is given by the formula:

F = qvB sinθ

Where:
F = Magnetic force
q = Charge of the particle
v = Velocity of the particle
B = Magnetic field strength
θ = Angle between the velocity and magnetic field

Analysis:
We need to find the change in radius of the circular path when the velocity of the charged particle is doubled and the strength of the magnetic field is halved.

1. Initial radius (r1): When the particle is moving with velocity v in the initial magnetic field B, it follows a circular path. The magnetic force provides the centripetal force required for the circular motion. Therefore, we can equate the magnetic force with the centripetal force:

F = qvB = mv²/r1

Where:
m = Mass of the particle

2. Final radius (r2): When the velocity of the charged particle is doubled (2v) and the strength of the magnetic field is halved (B/2), the magnetic force acting on the particle changes. We can equate the new magnetic force with the centripetal force:

F = q(2v)(B/2) = m(2v)²/r2

Solution:
By comparing the two equations, we can find the relationship between the initial and final radii:

qvB = m(2v)²/r1
q(2v)(B/2) = m(2v)²/r2

Simplifying the equations:

r1 = 2mv/qB
r2 = 4mv/qB

Taking the ratio of r2 and r1:

r2/r1 = (4mv/qB)/(2mv/qB)
r2/r1 = 4/2
r2/r1 = 2

Therefore, the radius becomes 2 times larger when the velocity of the charged particle is doubled and the strength of the magnetic field is halved. Hence, the correct answer is option 'c' - 2 times.

Answer

Magnetic Force on a Charged Particle
When a charged particle moves through a magnetic field, it experiences a magnetic force perpendicular to both the velocity of the particle and the magnetic field. This force is given by the equation:

F = qvBsinθ

Where:
F is the magnetic force
q is the charge of the particle
v is the velocity of the particle
B is the magnetic field strength
θ is the angle between the velocity vector and the magnetic field vector.

Effect of Doubling Velocity
If the velocity of the charged particle is doubled, the force experienced by the particle will also double. This is because the force is directly proportional to the velocity of the particle.

F' = 2F

Effect of Halving Magnetic Field Strength
If the magnetic field strength is halved, the force experienced by the particle will also be halved. This is because the force is directly proportional to the magnetic field strength.

F' = (1/2)F

Relationship between Force, Velocity, and Radius
The magnetic force on a charged particle moving in a magnetic field provides the necessary centripetal force to keep the particle moving in a circular path. The centripetal force is given by the equation:

F = mv^2/r

Where:
m is the mass of the particle
v is the velocity of the particle
r is the radius of the circular path.

Since the magnetic force (F) provides the necessary centripetal force for the circular motion, we can equate the two equations:

qvBsinθ = mv^2/r

Relationship between Radius and Velocity
From the equation relating force, velocity, and radius, we can rearrange the terms to solve for the radius:

r = mv/(qBsinθ)

Since the charge of the particle (q) and the angle between the velocity and magnetic field vectors (θ) are not changing, we can simplify the equation to:

r = mv/(qB)

Effect on Radius
When the velocity of the charged particle is doubled and the magnetic field strength is halved, the equation for the radius becomes:

r' = (2mv)/(q(1/2)B)
= 4(mv)/(qB)
= 4r

Therefore, the radius becomes 4 times its original value.

Answer

Explanation:

When a charged particle moves in a magnetic field, it experiences a force called the magnetic force. The magnitude of this force is given by the equation:

F = qvB sinθ

Where:
F is the magnetic force,
q is the charge of the particle,
v is the velocity of the particle,
B is the magnetic field strength, and
θ is the angle between the velocity vector and the magnetic field vector.

When a charged particle moves in a magnetic field with a velocity directed perpendicular to the magnetic field lines, it moves in a circular path. The centripetal force required to keep the particle moving in a circle is provided by the magnetic force:

F = mv^2/r

Where:
m is the mass of the particle, and
r is the radius of the circular path.

Initial Scenario:
In the initial scenario, the charged particle is moving with a velocity v and the magnetic field strength is B. Therefore, the magnetic force is:

F = qvB sinθ

Since the velocity is perpendicular to the magnetic field lines, the angle θ is 90 degrees and sinθ is equal to 1. Therefore, the magnetic force is:

F = qvB

The magnetic force provides the centripetal force required to keep the particle moving in a circle, so we can equate the two forces:

qvB = mv^2/r

Simplifying the equation, we get:

r = mv/qB

Final Scenario:
In the final scenario, the velocity of the charged particle is doubled (2v) and the strength of the magnetic field is halved (B/2). Therefore, the magnetic force is:

F = q(2v)(B/2) sinθ

Again, since the velocity is perpendicular to the magnetic field lines, the angle θ is 90 degrees and sinθ is equal to 1. Therefore, the magnetic force is:

F = qvB

The magnetic force provides the centripetal force required to keep the particle moving in a circle, so we can equate the two forces:

qvB = m(2v)^2/r'

Simplifying the equation, we get:

r' = m(2v)^2/qB

Comparing the equations for r and r', we can see that the only difference is the velocity term. Therefore, the ratio of r' to r is:

r'/r = (m(2v)^2/qB)/(mv/qB) = 4

Therefore, the radius becomes 4 times larger when the velocity is doubled and the magnetic field strength is halved. Thus, the correct answer is option 'B' - 4 times.

Answer

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