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Chapter 27: Problem 72

A positive charge is released from the origin at a place where uniform electric field \(E\) and a uniform magnetic field \(B\) exist along the positive \(y\) -axis and positive \(z\) -axis respectively, then: (a) initially the charge particle tends to move along positive \(z\) -axis (b) initially the charged particle tends to move along negative \(y\) -direction (c) initially the charged particle tends to move along positive \(y\) -direction (d) the charged particle moves in \(y-z\) plane

Short Answer

Expert verified
Initially, the charged particle tends to move along the positive \(y\)-direction.

Step by step solution

01

Determine the Electric Force

When a positive charge is placed in an electric field, it experiences a force in the direction of the electric field. Given that the electric field \(E\) is along the positive \(y\)-axis, the electric force \(F_{E}\) experienced by the charge is also in the positive \(y\)-direction. Thus, \(F_{E} = qE\), where \(q\) is the charge.
02

Determine the Magnetic Force

The magnetic force on a charged particle moving with velocity \(v\) in a magnetic field \(B\) is given by \(F_{B} = q(v \times B)\). Initially, since the charge is at rest, \(v = 0\), thus the magnetic force \(F_{B} = 0\). As the charge begins to move due to the electric field, only then will the magnetic field impact its motion.
03

Analyze Initial Motion

Since initially, the charge is at rest, the only acting force is the electric force along the positive \(y\)-axis. Thus, the charge will start moving in the direction of this electric force, which is the positive \(y\)-direction.
04

Consider Future Motion

Once the charge gains some velocity in the \(y\)-direction, the magnetic force will act perpendicular to both \(v\) and \(B\) (using right-hand rule, \(B\) is along \(z\)). This results in a cyclotron motion within the \(y-z\) plane.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Force
Electric force is the force exerted by an electric field on a charged particle. The direction of this force depends on the charge of the particle and the direction of the field. For a positive charge placed in an electric field, the force acts in the same direction as the field.

In the context of our exercise, the electric field is directed along the positive \(y\)-axis. Hence, if a positive charge is introduced, it experiences an electric force in the positive \(y\)-direction. This force can be calculated using the formula \(F_E = qE\), where \(q\) represents the charge and \(E\) is the strength of the electric field. This relationship is crucial as it sets the initial motion for the particle.

Understanding the electric force helps us predict the behavior of charged particles in fields. Once released, a positive charge will begin to move in the direction of this force unless other forces come into play.
Magnetic Force
Magnetic force is experienced by a charged particle when it moves through a magnetic field. The force arises because of the interaction between the particle鈥檚 velocity and the magnetic field. This is guided by the right-hand rule and is perpendicular to both the particle's velocity and the magnetic field.

Initially, if a particle is at rest in a magnetic field, the magnetic force is zero, since the force is determined by the cross product \(F_B = q(v \times B)\). Here, \(v\) is the velocity of the particle and \(B\) represents the magnetic field. If the particle is stationary, \(v = 0\), and therefore, \(F_B = 0\).

As a charged particle starts to move due to any other force, like an electric force, the magnetic field begins to exert its influence, altering the motion by acting at perpendicular angles to the instantaneous velocity and field directions.
Cyclotron Motion
Cyclotron motion describes the circular path taken by a charged particle under the influence of perpendicular electric and magnetic forces. Once a charged particle in an electric field gains velocity, a magnetic field can influence its path, giving it a circular or spiral trajectory.

In our scenario, once the particle gains some velocity down the \(y\)-axis due to the electric force, a magnetic force becomes relevant. Given that the magnetic field is along the \(z\)-axis, the particle鈥檚 path curves in the \(y-z\) plane, leading to cyclotron motion. This is due to the cross-product nature of the magnetic force, acting at right angles to both motion and field directions.

This motion is fundamental in devices like cyclotrons, which accelerate charged particles in a spiral path, demonstrating how combined electric and magnetic fields can be harnessed to control particle trajectories effectively. Understanding cyclotron motion helps explain numerous phenomena in particle physics and electromagnetism.

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Most popular questions from this chapter

Through two parallel wires \(P\) and \(Q .10 \mathrm{~A}\) and \(2 \mathrm{~A}\) of currents are passed respectively in opposite directions. If the wire \(P\) is infinitely long and the length of the wire \(Q\) is \(2 \mathrm{~cm}\), the force on the conductor \(Q\) which is situated at \(10 \mathrm{~cm}\) distance from \(P\), will be : (a) \(8 \times 10^{-5} \mathrm{~N}\) (b) \(3 \times 10^{-7} \mathrm{~N}\) (c) \(4 \times 10^{-5} \mathrm{~N}\) (d) \(4 \times 10^{-7} \mathrm{~N}\)

An electron accelerated through a potential difference \(V\) enters into a uniform transverse magnetic field and experiences a force \(F\). If the accelerating potential is increased to \(2 V\), the electron in the same magnetic field will experience a force: (a) \(\bar{F}\) (b) \(\frac{F}{2}\) (c) \(\sqrt{2} F\) (d) \(2 F\)

A conducting circular loop of radius \(r\) carries a constant current \(I\). It is placed in a uniform magnetic field \(B\) such that \(B_{0}\) is perpendicular to the plane of the loop. The magnetic force acting on the loop is: (a) \(\operatorname{Ir} B_{0}\) (b) \(2 \pi \operatorname{lr} B_{0}\) (c) \(\pi I r B_{0}\) (d) zero

A charged particle of mass \(m\) and charge \(q\) in a uniform magnetic field \(B\) acts into the plane. The plane is frictional having coefficient of friction \(\mu\). The speed of charged particle just before entering into the region is \(v_{0}\). The radius of curvature of the path after the time \(\frac{v_{0}}{2 \mu g}\) is : (a) \(\frac{m v_{0}}{q B}\) (b) \(\frac{m v_{0}}{2 q B}\) (c) \(\frac{m v_{0}}{4 q B}\) (d) none of these

A charged particle enters into a uniform magnetic field \(B\) with velocity \(\vec{v}\) at an angle \(\theta\) as shown in figure. Then the ratio of radius to pitch of helix is: (a) \(\frac{2 \pi}{\tan \theta}\) (b) \(\tan \theta\) (c) \(\cot \theta\) (d) \(\frac{\tan \theta}{2 \pi}\)
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