Answered: Find the direction and magnitude of the…

College Physics

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ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter1: Introduction: The Nature Of Science And Physics

Section: Chapter Questions

Problem 22PE: What is the area of a circle 3.102 cm in diameter?

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Question

Find the direction and magnitude of the minimum velocity of charge q = +2e that will produce the magnetic force F shown in the figure below. (Measure the angle counterclockwise from the x axis in the xz plane. Take the magnitudes of F as 1.96 x 10-11 N and the magnitude of B as 2.19 T.)

magnitude
direction

Expert Solution

Step 1

Given:

${Force, F = 1.96×10}^{- 11} N Magnetic field, B = 2.19 T Charge, q = + 2 e = 2 \times (1.6 \times 10^{- 19} C) = + 3.2 \times 10^{- 19} C$

Solution:

We know that, the force F is perpendicular to the velocity 'v' and magnetic force 'B'.

Let us consider velocity is perpendicular to magnetic field $\vec{B}$ and electric field $\vec{E}$ , therefore velocity 'v' is $45 °$ counter-clockwise from x-axis in x-z plane and the direction of particle is along : $(x \hat{i} - z \hat{j})$

Consider the following equation:

$F = q v B where, q is the charge v is velocity and B is magnetic field$

Substituting the given values in above equation, we get:

$\begin{array}{rcl} F & = & q v B \\ (1.96 \times 10^{- 11} N) & = & (3.2 \times 10^{- 19} C) \times (v) \times (2.19 T) \end{array}$

Re-arranging the above equation:

$\begin{array}{rcl} v & = & \frac{(1.96 \times 10^{- 11} N)}{(3.2 \times 10^{- 19}) \times (2.19 T)} \\ = & 2.8 \times 10^{7} m / s \end{array}$

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