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Chapter 15: Problem 140

An electron accelerated by a potential difference \(V=3.6 \mathrm{~V}\) first enters into a uniform electric field of a parallel-plate capacitor whose plates extend over a length \(l=6 \mathrm{~cm}\) in the direction of initial velocity. The electric field is normal to the direction of initial velocity and its strength varies with time as \(E=a \times t\), where \(a=\) \(3200 \mathrm{Vm}^{-1} \mathrm{~s}^{-1}\). Then the electron enters into a uniform magnetic field of induction \(B=\pi \times 10^{-9} \mathrm{~T}\). Direction of magnetic field is same as that of the electric field. Calculate pitch (in \(\mathrm{mm}\) ) of helical path traced by the electron in the magnetic field (Mass of electron, \(m=9 \times\) \(10^{-31} \mathrm{~kg}\) ). [Neglect the effect of induced magnetic field.]

Short Answer

Expert verified
The pitch (p) of the helical path traced by the electron in the magnetic field is approximately \(7.66 \mathrm{~mm}\).

Step by step solution

01

Determine the acceleration of the electron in the electric field

First, let's convert the given electric field formula from variable strength E(t) to acceleration formula a_y(t): a_y(t) = e * E(t) / m where e is the electron charge, m is the electron mass, and E(t) = a * t. Now we can find acceleration a_y as a function of time using the given parameter values: a_y(t) = \( (-e)(3200 t)/(9\times10^{-31}kg)\).
02

Calculate initial velocity of electron

Use the potential difference (V) to determine the initial kinetic energy (K) and velocity (v_x) of the electron. K = e * V v_x = \(\sqrt{(2K)/m}\) Plug in values: v_x = \(\sqrt{(2(-e)(3.6 V))/(9 \times 10^{-31} kg)}\)
03

Position of electron in electric field

Use the time-based position formula for electron in electric field to determine y-coordinate: y(t) = (1/2) a_y(t) * (t^2) To find the time when the electron exits the electric field, use the length of plates: l = v_x * t_exit Solve for t_exit and calculate y(t_exit).
04

Determine time period for helical motion

Time period of helical motion in magnetic field (T) can be calculated as: T = \(2\pi m/(eB)\) Plug in the given values for B and m, and calculate T.
05

Calculate the pitch of the helical path

The pitch (p) can be calculated by multiplying the vertical velocity (v_y) at the end of the first region with the time period (T) of one complete helical motion. v_y = a_y(t_exit) * t_exit And then find the pitch: p = v_y * T Use the calculated values to determine the pitch of the electron's path in the magnetic field. Then, convert the pitch (p) from meters to millimeters to obtain the final answer.

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Most popular questions from this chapter

Two infinitely long, thin, insulated, straight wires lies in the \(x-y\) plane along the \(x\) - and \(y\)-axes, respectively. Each wire carries a current \(I\), respectively, in the positive \(x\)-direction and positive \(y\)-direction. The magnetic field will be zero at all points on the straight line. (A) \(y=x\) (B) \(y=-x\) (C) \(y=x-1\) (D) \(y=-x+1\)

A uniform horizontal magnetic field \(\vec{B}\) exists in the region. A rectangular loop of mass \(m\), horizontal side \(l\) (perpendicular to the magnetic field), and resistance \(R\) is placed in the region. The velocity with which it should be pushed down so that it continues to fall without acceleration will be (A) \(\frac{m g R}{B^{2} l^{2}}\) (B) \(\frac{B^{2} l^{2}}{m g R}\) (C) \(\frac{m g}{B l R}\) (D) \(\frac{R B l}{m g}\)

A circular coil \(A\) has a radius \(R\) and the current flowing through it is \(I\). Another circular coil \(B\) has radius \(2 R\) and if \(2 I\) is the current flowing through it, then the magnetic field at the centre of the circular coil are in the ratio of (A) \(4: 1\) (B) \(2: 1\) (C) \(3: 1\) (D) \(1: 1\)

Two short bar magnets of magnetic moments \(M\) each are arranged at the opposite corners of a square of side \(d\) such that their centres coincide with the corners and their axes are parallel. If the like poles are in the same direction, the magnetic induction at any of the other corners of the square is (A) \(\frac{\mu_{0}}{4 \pi} \frac{M}{d^{3}}\) (B) \(\frac{\mu_{0}}{4 \pi} \frac{2 M}{d^{3}}\) (C) \(\frac{\mu_{0}}{4 \pi} \frac{M}{2 d^{3}}\) (D) \(\frac{\mu_{0}}{4 \pi} \frac{M^{2}}{2 d^{3}}\)

The distance between two thin long straight parallel conducting wires is \(b\). On passing the same current \(i\) in them, the force per unit length between them will be (A) Zero (B) \(\frac{\mu_{0} i^{2}}{2 \pi b}\) (C) \(\frac{\mu_{0} i}{2 \pi b}\) (D) \(\frac{\mu_{0} i}{2 \pi}\)
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