Chapter 20, Problem 9Q

To determine

The direct on of the magnetic force due to the magnetic field produced

by the wire for each particle.

Answer to Problem 9Q

Solution:

Direction of magnetic force in each case are shown in figure.

Explanation of Solution

Introduction:

Direction of magnetic field due to current carrying wire is given by right hand curl rule.

Right hand rule says that extend your right-hand thumb in the direction of current then direction of curl of your four fingers of right hand gives the direction of magnetic field.

Magnetic force $\left({\overrightarrow{F}}_m\right)$ is given by

${\overrightarrow{F}}_m==q\left(\overrightarrow{V}\times \overrightarrow{B}\right)$

Hence direction of magnetic field is given by vector product.

(a)For a:

Direction of $\overrightarrow{V}\text{ is }\stackrel{\wedge }{i}$

Direction of $\overrightarrow{B}\text{ is }\stackrel{\wedge }{K}$

So direction of ${\overrightarrow{F}}_m$ is given by the direction of $q\left(\overrightarrow{V}\times \overrightarrow{B}\right)$

Direction of ${\overrightarrow{F}}_m$ is given by $\left(\stackrel{\wedge }{i}\times \stackrel{\wedge }{K}\right)$

$=-\stackrel{\wedge }{J}$

Hence magnetic field is in $-Ve$ y direction.

(b)For b

Direction of $\overrightarrow{V}\text{ is }\stackrel{\wedge }{i}$

Direction of $\overrightarrow{B}\text{ is }\stackrel{\wedge }{K}$

So direction of ${\overrightarrow{F}}_m$ is given by $-Ve$ of direction of direction of $\left(\overrightarrow{V}\times \overrightarrow{B}\right)$.

Direction of ${\overrightarrow{F}}_m=\text{direction of }-\left(\overrightarrow{V}\times \overrightarrow{B}\right)$

$\begin{array}{l}=\text{direction }-\left(\stackrel{\wedge }{J}\times \stackrel{\wedge }{K}\right)\\ =-\stackrel{\wedge }{i}\end{array}$

${\overrightarrow{F}}_m$ Is acting towards $-Ve$ x-axis as shown in figure.

(c)For C:

Direction of $\overrightarrow{V}\text{ is }-\stackrel{\wedge }{i}$

Direction of $\overrightarrow{B}\text{ is }-\stackrel{\wedge }{K}$

So direction of ${\overrightarrow{F}}_m$ is given by direction of $q\left(\overrightarrow{V}\times \overrightarrow{B}\right)$. Here charge is $-Ve$.

Direction of ${\overrightarrow{F}}_m=\text{direction of }-\left(\overrightarrow{V}\times \overrightarrow{B}\right)$

($-Ve$ Sign is take for $-Ve$ charge)

Direction of ${\overrightarrow{F}}_m=\text{direction of }-\left(-\stackrel{\wedge }{i}\times -\stackrel{\wedge }{K}\right)$

$\begin{array}{l}=-\left(\stackrel{\wedge }{i}\times \stackrel{\wedge }{K}\right)\\ =-\left(-\stackrel{\wedge }{J}\right)\left(\stackrel{\wedge }{i}\times \stackrel{\wedge }{K}=-\stackrel{\wedge }{J}\right)\\ =\stackrel{\wedge }{J}\end{array}$

Hence magnetic force active towards the $+Ve$ y axis as shown in figure.

(d)For d:

Direction of $\overrightarrow{V}=\stackrel{\wedge }{J}$

Direction of $\overrightarrow{B}\text{ is }-\stackrel{\wedge }{K}$

So direction of ${\overrightarrow{F}}_m$ is given by the direction of $q\left(\overrightarrow{V}\times \overrightarrow{B}\right)$

Direction of $q\left(\overrightarrow{V}\times \overrightarrow{B}\right)$ is $+\left(\stackrel{\wedge }{J}\times \stackrel{\wedge }{K}\right)$

$\left(+\text{ sign for +Ve charge}\right)$

$\begin{array}{l}=\left(\stackrel{\wedge }{J}\times \stackrel{\wedge }{K}\right)\\ =\stackrel{\wedge }{i}\end{array}$

Hence magnetic force is acting towards the $-Ve$ x axis as shown in figure.