Chapter 4: Q8E (page 912)

A particle with charge $- 5.60 nC$ is moving in a uniform magnetic fieldrole="math" localid="1655717557369" $B = - (1.25 T) \hat{k}$
The magnetic force on the particle is measured to berole="math" localid="1655717706597" $\overset{⇀}{F} = - (3.40 \times 10^{- 7} N) \hat{i} - (7.40 \times 10^{- 7} N) \hat{j}$ (a) Calculate all the components of the velocity of the particle that you can from this information. (b) Are there
components of the velocity that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product $\overset{⇀}{v} ֏ \overset{⇀}{F}$ . What is the angle between velocity and force?

Short Answer

Expert verified

(a) The velocity along x axis is $V_{x} = - 106 m / s$ and along y axis is $V_{y} = - 48.6 m l s$

(b) The z component is not determined, since there is no force along this component hence measurement doesn’t tell us anything about $V_{z}$

Step by step solution

The significance of magnetic field

The magnetic field force is given by

$F_{B} = q (v \times B) F = q v B \sin θ$

Where q is the charge of the particle, V is the velocity and B is the magnetic field

Determination of the components of the velocity of the particle

We know that the magnetic field force is given by

$F = q (v \times B)$

We know

$\overset{⇀}{F} = q ֏ d e t (\begin{matrix} i & j & k \\ v_{x} & v_{y} & v_{z} \\ B_{x} & B_{y} & B_{z} \end{matrix})$

role="math" localid="1655718671037"

F_{x} \hat{i} + F_{y} \hat{j} + F_{z} \hat{k} = q (v_{y} B_{z} - v_{z} B_{y}) \hat{i} + q (v_{z} B_{x} - v_{x} B_{z}) \hat{j} + q (v_{x} B_{y} - v_{y} B_{x}) \hat{k}

We know that $B_{y} = B_{x} = 0$

Applying this in the equation we get

$F_{x} \hat{i} + F_{y} \hat{j} + F_{z} \hat{k} = q v_{y} B_{z} \hat{i} - q v_{x} B_{z} \hat{j}$

Equating the two sides we get

$F_{x} = q v_{y} B_{z} F_{y} = - q v_{x} B_{z} F_{z} = 0$

Plugging in the values we get

$v_{x} = \frac{F_{y}}{- q B_{z}} = \frac{7.40 \times 10^{- 7} N}{- (5.60 \times 10^{- 9} C) (- 1.25 T)} = - 106 m / s V_{y} = \frac{F_{x}}{q B_{z}} = \frac{- 3.40 \times 10^{- 7} N}{(5.60 \times 10^{- 9} C) (- 1.25 T)} = - 48.6 m / s$

Hence, the velocity along x axis is $v_{x} = - 106 m / s$ and along y axis is $v_{y} = - 48.6 m / s$

Determination scalar Product

We see

$\overset{⇀}{v} ֏ \overset{⇀}{F} = v_{x} F_{x} + V_{y} F_{y} + V_{z} F_{z} \overset{⇀}{v} ֏ \overset{⇀}{F} = \frac{F_{y}}{- q B_{z}} F_{x} + \frac{F_{x}}{q B_{z}} F_{y} \overset{⇀}{v} ֏ \overset{⇀}{F} = 0$