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A uniform magnetic field in z direction is exist as shown a charged particle of mass m and charge Q greater than zero is released at the origin in the y z plane with a velocity v directed at an angle theta equals to 45 degree with respect to positive z Axis ignoring gravity which of the following is true the particle moves in circular path or the particle continuous in a straight line with constant speed all the initial acceleration of particle is a equals to qv not by √2m in positive X direction?
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A uniform magnetic field in z direction is exist as shown a charged pa...
Magnetic Field and Charged Particle Motion
When a charged particle is placed in a magnetic field, its motion is influenced by the Lorentz force. Here’s an analysis of the situation described:
Initial Conditions
- The charged particle has mass \( m \) and charge \( Q \) (where \( Q > 0 \)).
- It is released from the origin in the y-z plane with an initial velocity \( \mathbf{v} \).
- The velocity makes an angle \( \theta = 45^\circ \) with the positive z-axis.
Understanding the Motion
- The velocity can be decomposed into components:
- \( v_x = v \cdot \sin(45^\circ) = \frac{v}{\sqrt{2}} \)
- \( v_y = 0 \)
- \( v_z = v \cdot \cos(45^\circ) = \frac{v}{\sqrt{2}} \)
Force Acting on the Particle
- The particle experiences a magnetic force given by:
\[
\mathbf{F} = Q(\mathbf{v} \times \mathbf{B})
\]
- Assuming the magnetic field \( \mathbf{B} \) is in the z-direction, the force will act in the x-y plane.
Resulting Motion
- Since the magnetic force is always perpendicular to the velocity, it causes the particle to undergo circular motion in the x-y plane.
- The initial acceleration can be calculated using:
\[
a = \frac{QvB}{m}
\]
(where \( B \) is the magnitude of the magnetic field).
Conclusion
- The particle does not move in a straight line; instead, it follows a circular path in the x-y plane due to the magnetic force acting on it.
- The initial acceleration in the x-direction is correctly identified as:
\[
a = \frac{Qv}{\sqrt{2}m}
\]
indicating that the particle will spiral away from the origin due to the continuous change in direction of the velocity vector.
This analysis confirms that the presence of a magnetic field results in circular motion for the charged particle.
When a charged particle is placed in a magnetic field, its motion is influenced by the Lorentz force. Here’s an analysis of the situation described:
Initial Conditions
- The charged particle has mass \( m \) and charge \( Q \) (where \( Q > 0 \)).
- It is released from the origin in the y-z plane with an initial velocity \( \mathbf{v} \).
- The velocity makes an angle \( \theta = 45^\circ \) with the positive z-axis.
Understanding the Motion
- The velocity can be decomposed into components:
- \( v_x = v \cdot \sin(45^\circ) = \frac{v}{\sqrt{2}} \)
- \( v_y = 0 \)
- \( v_z = v \cdot \cos(45^\circ) = \frac{v}{\sqrt{2}} \)
Force Acting on the Particle
- The particle experiences a magnetic force given by:
\[
\mathbf{F} = Q(\mathbf{v} \times \mathbf{B})
\]
- Assuming the magnetic field \( \mathbf{B} \) is in the z-direction, the force will act in the x-y plane.
Resulting Motion
- Since the magnetic force is always perpendicular to the velocity, it causes the particle to undergo circular motion in the x-y plane.
- The initial acceleration can be calculated using:
\[
a = \frac{QvB}{m}
\]
(where \( B \) is the magnitude of the magnetic field).
Conclusion
- The particle does not move in a straight line; instead, it follows a circular path in the x-y plane due to the magnetic force acting on it.
- The initial acceleration in the x-direction is correctly identified as:
\[
a = \frac{Qv}{\sqrt{2}m}
\]
indicating that the particle will spiral away from the origin due to the continuous change in direction of the velocity vector.
This analysis confirms that the presence of a magnetic field results in circular motion for the charged particle.
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A uniform magnetic field in z direction is exist as shown a charged particle of mass m and charge Q greater than zero is released at the origin in the y z plane with a velocity v directed at an angle theta equals to 45 degree with respect to positive z Axis ignoring gravity which of the following is true the particle moves in circular path or the particle continuous in a straight line with constant speed all the initial acceleration of particle is a equals to qv not by √2m in positive X direction?
Question Description
A uniform magnetic field in z direction is exist as shown a charged particle of mass m and charge Q greater than zero is released at the origin in the y z plane with a velocity v directed at an angle theta equals to 45 degree with respect to positive z Axis ignoring gravity which of the following is true the particle moves in circular path or the particle continuous in a straight line with constant speed all the initial acceleration of particle is a equals to qv not by √2m in positive X direction? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A uniform magnetic field in z direction is exist as shown a charged particle of mass m and charge Q greater than zero is released at the origin in the y z plane with a velocity v directed at an angle theta equals to 45 degree with respect to positive z Axis ignoring gravity which of the following is true the particle moves in circular path or the particle continuous in a straight line with constant speed all the initial acceleration of particle is a equals to qv not by √2m in positive X direction? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniform magnetic field in z direction is exist as shown a charged particle of mass m and charge Q greater than zero is released at the origin in the y z plane with a velocity v directed at an angle theta equals to 45 degree with respect to positive z Axis ignoring gravity which of the following is true the particle moves in circular path or the particle continuous in a straight line with constant speed all the initial acceleration of particle is a equals to qv not by √2m in positive X direction?.
A uniform magnetic field in z direction is exist as shown a charged particle of mass m and charge Q greater than zero is released at the origin in the y z plane with a velocity v directed at an angle theta equals to 45 degree with respect to positive z Axis ignoring gravity which of the following is true the particle moves in circular path or the particle continuous in a straight line with constant speed all the initial acceleration of particle is a equals to qv not by √2m in positive X direction? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A uniform magnetic field in z direction is exist as shown a charged particle of mass m and charge Q greater than zero is released at the origin in the y z plane with a velocity v directed at an angle theta equals to 45 degree with respect to positive z Axis ignoring gravity which of the following is true the particle moves in circular path or the particle continuous in a straight line with constant speed all the initial acceleration of particle is a equals to qv not by √2m in positive X direction? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniform magnetic field in z direction is exist as shown a charged particle of mass m and charge Q greater than zero is released at the origin in the y z plane with a velocity v directed at an angle theta equals to 45 degree with respect to positive z Axis ignoring gravity which of the following is true the particle moves in circular path or the particle continuous in a straight line with constant speed all the initial acceleration of particle is a equals to qv not by √2m in positive X direction?.
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