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Chapter 19: Problem 16

Current \(i_{0}\) is passes through a solenoid of length \(l\) having number of turns \(N\) when it is connected to a DC source. A charged particle with charge \(q\) is projected along the axis of the solenoid with a speed \(v_{0} .\) The velocity of the particle in the solenoid (a) increases (b) decreases (c) remain same (d) becomes zero

Short Answer

Expert verified
(c) The velocity of the particle remains the same.

Step by step solution

01

Determine the magnetic field inside the solenoid

Inside a solenoid, the magnetic field is uniform and given by the formula \( B = \mu_0 \frac{N}{l} i_0 \), where \( \mu_0 \) is the permeability of free space, \( N \) is the number of turns, \( l \) is the length, and \( i_0 \) is the current passing through the solenoid.
02

Analyze motion of the charged particle

The charged particle with charge \( q \) is projected along the axis of the solenoid. Upon entering a uniform magnetic field, the force on the particle is given by \( F = q(\vec{v} \times \vec{B}) \). Since the particle moves along the solenoid axis, \( \vec{v} \) is parallel to \( \vec{B} \), making the cross product zero.
03

Evaluate the velocity change

Since the magnetic force \( F = q(\vec{v} \times \vec{B}) \) is zero, there is no force acting on the particle to change its speed. Hence, the velocity of the particle does not change.

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Key Concepts

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

Charged Particle Motion
Understanding how a charged particle moves in a magnetic field is essential in physics. A charged particle, such as an electron or proton, experiences forces that can change its motion. When a charged particle is introduced into a magnetic field, its path and speed can be influenced. However, this is highly dependent on the orientation of the particle's velocity relative to the magnetic field.

- If the particle鈥檚 velocity vector is perpendicular to the magnetic field, it experiences a force that causes it to move in a circular path. - If the velocity is parallel to the field, as in our exercise with the solenoid, the charged particle travels straight along the path without deviation.

These principles apply when the magnetic field is uniform, an environment you commonly find in devices like solenoids.
Uniform Magnetic Field
A uniform magnetic field possesses the same magnitude and direction at every point. Solenoids are often used to create such fields efficiently. Inside a solenoid, the field lines are parallel, equally spaced, and closely grouped, resulting in uniform strength along its axis.

- The magnetic field inside a solenoid is calculated using the formula: \( B = \mu_0 \frac{N}{l} i_0 \).- The symbols \( \mu_0 \) indicate the permeability of free space, \( N \) is the number of coil turns, \( l \) is the length of the solenoid, and \( i_0 \) represents the current flowing through it.

Understanding uniform magnetic fields helps in predicting how charged particles will behave when they're introduced into such a field, as it is predictable and consistent in strength and direction. This predictability is why solenoids are utilized in numerous applications, from electromagnets to inductive sensors.
Lorentz Force
The Lorentz Force is a fundamental concept that explains how charged particles move in magnetic and electric fields. It combines electrical and magnetic forces acting upon charged particles. Specifically, the magnetic component of the Lorentz Force is critical when discussing motion in a solenoid.

- The magnetic force part of the Lorentz Force is described by the equation \( F = q(\vec{v} \times \vec{B}) \), where \( F \) is force, \( q \) is electric charge, \( \vec{v} \) is the particle's velocity, and \( \vec{B} \) is the magnetic field.- In situations like our solenoid exercise, if the velocity \( \vec{v} \) of the particle is parallel to the magnetic field \( \vec{B} \), then the cross product results in zero force.

Consequently, when there's no force acting on a charged particle due to the parallel nature of velocity and magnetic field, there is no change in speed. Therefore, the particle maintains its velocity as it passes through the solenoid.

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Most popular questions from this chapter

A long straight, solid metal wire of radius \(2 \mathrm{~mm}\) carries a current uniformly distributed over its circular cross-section. The magnetic field induction at a distance \(2 \mathrm{~mm}\) from its axis is \(B\). Then, the magnetic field induction at distance \(1 \mathrm{~mm}\) from axis will be (a) \(B\) (b) \(B_{1} 2\) (c) \(2 \underline{B}\) (d) \(B\)

The magnetic field normal to the plane of a wire of \(n\) turns and radius \(r\) which carries a current \(i\) is measured on the axis of the coil at a small distance \(h\) from the centre of the coil. This is smaller than the magnetic field at the centre by the fraction (a) \((2 / 3) r^{2} / h^{2}\) (b) \((3 / 2) r^{2} / h^{2}\) (c) \((2 / 3) h^{2} / r^{2}\) (d) \((3 / 2) h^{2} / r^{2}\)

Consider a wire carrying a steady current, \(I\) placed in a uniform magnetic field \(\mathbf{B}\) perpendicular to its length. Consider the charges inside the wire. It is known that magnetic forces do no work. This implies that, \(\quad\) [NCERT Exemplar] (a) motion of charges inside the conductor is unaffected by B since they do not absorb energy (b) some charges inside the wire move to the surface as a result of B (c) if the wire moves under the influence of \(\mathrm{B}\), no work is done by the force (d) if the wire moves under the influence of \(\mathrm{B}\), no work is done by the magnetic force on the ions, assumed fixed within the wire

Two very long straight parallel wires carry currents \(i\) and \(2 i\) in opposite directions. The distance between the wires is \(r\). At a certain instant of time a point charge, \(q\) is at a point equidistant from the two wires in the plane of the wires. Its instantaneous velocity \(v\) is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is (a) zero (b) \(\frac{3 \mu_{0}}{2 \pi} \frac{i q v}{r}\) (c) \(\frac{\mu_{0}}{\pi} \frac{i q v}{r}\) (d) \(\frac{\mu_{0}}{2 \pi} \frac{i q v}{r}\)

An element, \(d l=d x\) \hat{ i } (where \(d x=1 \mathrm{~cm}\) ) is placed at the origin and carries a large current \(i=10 \mathrm{~A}\). What is the magnetic field on the \(y\)-axis at a distance of \(0.5 \mathrm{~m} ?\) (a) \(2 \times 10^{-8} \hat{\mathrm{k}} \mathrm{T}\) (b) \(4 \times 10^{-8} \hat{\mathrm{k}} \mathrm{T}\) (c) \(-2 \times 10^{-8} \hat{\mathrm{k}} \mathrm{T}\) (d) \(-4 \times 10^{-8} \hat{\mathrm{k}} \mathrm{T}\)
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