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Chapter 13: Problem 1911

1: A charged particle moves in a uniform mag. field. The velocity of the particle at some instant makes an acute angle with the mag. field. The path of the particle will be (a) A straight line (b) A circle (c) A helix with uniform pitch (d) A helix with non-uniform pitch

Short Answer

Expert verified
The path of the charged particle will be a \(helix\) with a uniform pitch when its velocity makes an acute angle with the magnetic field. This is due to the combination of motion along the magnetic field (straight line) and the motion perpendicular to the magnetic field (circle) caused by the Lorentz force acting on the particle.

Step by step solution

01

Identify the force acting on the charged particle

: When a charged particle with charge q moves with a velocity \(\vec{v}\) in a magnetic field \(\vec{B}\), it experiences a force called the Lorentz force. This force can be calculated using the following formula: \( \vec{F} = q( \vec{v} \times \vec{B} ) \) where \(\vec{F}\) is the Lorentz force and \(\times\) denotes the cross product.
02

Analyze the motion of the particle in the magnetic field

: The force acting on the particle is always perpendicular to its velocity (\(\vec{v}\)). Therefore, when the charged particle enters the magnetic field at an acute angle, the trajectory can be divided into two components. The first component is along the magnetic field and the second component is perpendicular to the magnetic field.
03

Determine the path along the magnetic field

: Since the force acting on the particle is always perpendicular to its velocity, there will be no force acting on a particle moving in the direction of the magnetic field. Thus, along the magnetic field, the particle will continue moving with a constant velocity, which means that it would form a straight line along the magnetic field direction.
04

Determine the path perpendicular to the magnetic field

: Perpendicular to the magnetic field, the charged particle experiences a centripetal force due to the Lorentz force, which causes the particle to move in a circular path. The radius of this circle can be determined by setting the centripetal force equal to the magnetic force: \( \frac{mv^2}{r} = qvB \) where m is the mass of the particle, v is its velocity perpendicular to the magnetic field, and r is the radius of the circular path.
05

Combine both paths to find the complete path

: Since the motion of the particle is a combination of the motion along the magnetic field (straight line) and the motion perpendicular to the magnetic field (circle), the complete path of the charged particle will be a helix with a uniform pitch. Thus, the correct option is: (c) A helix with uniform pitch

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Most popular questions from this chapter

A straight wire of mass \(200 \mathrm{gm}\) and length \(1.5\) meter carries a current of 2 Amp. It is suspended in mid-air by a uniform horizontal magnetic field B. [take \(\left.\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right]\). The \(\mathrm{B}\) is (a) \(\overline{(2 / 3) \text { tes } 1 a}\) (b) \((3 / 2)\) tesla (c) \((20 / 3)\) tesla (d) (3/20) tesla

A long horizontal wire " \(\mathrm{A}^{\prime \prime}\) carries a current of $50 \mathrm{Amp}$. It is rigidly fixed. Another small wire "B" is placed just above and parallel to " \(\mathrm{A}^{\prime \prime}\). The weight of wire- \(\mathrm{B}\) per unit length is \(75 \times 10^{-3}\) Newton/meter and carries a current of 25 Amp. Find the position of wire \(B\) from \(A\) so that wire \(B\) remains suspended due to magnetic repulsion. Also indicate the direction of current in \(B\) w.r.t. to \(A\).(a) \((1 / 2) \times 10^{-2} \mathrm{~m}\); in same direction (b) \((1 / 3) \times 10^{-2} \mathrm{~m}\); in mutually opposite direction (c) \((1 / 4) \times 10^{-2} \mathrm{~m}\); in same direction (d) \((1 / 5) \times 10^{-2} \mathrm{~m}\); in mutually opposite direction

In a mass spectrometer used for measuring the masses of ions, the ions are initially accelerated by an ele. potential \(\mathrm{V}\) and then made to describe semicircular paths of radius \(\mathrm{r}\) using a magnetic field \(\mathrm{B}\). If \(\mathrm{V}\) and \(\mathrm{B}\) are kept constant, the ratio [(Charge on the ion) / (mass of the ion)] will be proportional to. (a) \(\left(1 / r^{2}\right)\) (b) \(r^{2}\) (c) \(\mathrm{r}\) (d) \((1 / \mathrm{r})\)

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A Galvanometer has a resistance \(\mathrm{G}\) and \(\mathrm{Q}\) current \(\mathrm{I}_{\mathrm{G}}\) flowing in it produces full scale deflection. \(\mathrm{S}_{1}\) is the value of the shunt which converts it into an ammeter of range 0 to \(\mathrm{I}\) and \(\mathrm{S}_{2}\) is the value of the shunt for the range 0 to \(2 \mathrm{I}\). The ratio $\left(\mathrm{S}_{1} / \mathrm{S}_{2}\right) \mathrm{is}$ (a) $\left[\left(2 \mathrm{I}-\mathrm{I}_{\mathrm{G}}\right) /\left(\mathrm{I}-\mathrm{I}_{\mathrm{G}}\right)\right]$ (b) $(1 / 2)\left[\left(\mathrm{I}-\mathrm{I}_{\mathrm{G}}\right) /\left(2 \mathrm{I}-\mathrm{I}_{\mathrm{G}}\right)\right]$ (c) 2 (d) 1
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