Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 51

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Weekly Quiz Competition

Question 1
1 / -0

Electron of mass m and charge q is travelling with a speed v along a circular path of radius r at right angles to a uniform magnetic field of intensity B. If the speed of the electron is doubled and the magnetic field is halved the resulting path would have a radius.

A
$2r$

B
$4r$

C
$r/4$

D
$r/2$

Solution

Given,
Mass of electron=m
Charge of electron=q
Velocity of electron=V
Radius of electron = r
Magnetic field =B
We know that force due to magnetic field, $F_B=qvB$
Now if $V^{'}=2V\text { and } B^{'}=\dfrac{B}{2}$
Then $F_B^{'}=qV^{'}B^{'}=q(2V)(\dfrac{B}{2})$
$F_B=\dfrac{mV^2}{r}$
$\implies r=\dfrac{mV^2}{qVB}\implies r=\dfrac{mV}{qB}$ and
${ r }^{ ' }=\dfrac { m\left( 2V \right) }{ q\left( \dfrac { B }{ 2 } \right) }$
$r^{'}=4r$
Question 2
1 / -0

Three long straight wires A, B and C are carrying currents as shown in figure. Then the resultant force on B is directed :

A
perpendicular to the plane of the paper and outward

B
perpendicular to the plane of the paper and inward

C
towards A

D
towards C

Solution

Hint:
If the current in the two parallel wires is in the same direction then the wires attract each other.

Step 1: Expression of force between two parallel wires,
The force between two parallel wires having current $I_1$ and $I_2$ and separated by distance $d$ is,
$F=\frac{\mu_oI_1 I_2}{2\pi d}$

Therefore,
$F\propto I_1I_2$ and $F\propto \frac{1}{d}$

Step 2: Find the stronger force.
Wire A and wire C are trying to pull wire B towards them.
The separation between the wires for both cases is the same.
$I_A I_B=(1A)(2A)=3A^2$
$I_c I_B=(3A)(2A)=6A^2$
$\because$ $F\propto I_1I_2$
$\therefore$ $F_{BC}>F_{AB}$

Therefore wire is pulled towards $C$ .
$\textbf{Hence option D correct}$
Question 3
1 / -0

A long curved conductor carrier a current $\vec{I}$ ( $\vec{I}$ is a vector). A small current element of length $d \vec l$ , on the wire, induces a magnetic field at a point, away from the current element. If the position vector between the current element and the point is $\vec{r}$ , making an angle $\theta$ with current element then, the induced magnetic field density, $d \vec B$ at the point is $(\mu_0=$ permeability of free space):

A
$\displaystyle\frac{\mu_oI (d \vec l\times \vec{r})}{4\pi r^3}$ perpendicular to the current element $d \vec l$

B
$\displaystyle\frac{\mu_oI\times \vec{r}.d \vec l}{4\pi r^2}$ perpendicular to the current element $d \vec l$

C
$\displaystyle\frac{\mu_oI\times d \vec l}{ r}$ perpendicular to the plane containing the current element and position vector $\vec{r}$

D
$\displaystyle\frac{\mu_oI\times d \vec l}{4\pi r^2}$ perpendicular to the plane containing current element and position vector $\vec{r}$

Solution

Biot Savart law provides the general expression for magnetic field induced by a small current carrying element at a distance from the element in a given direction. It states,

$d \vec B=\dfrac { { \mu }_{ 0 }I(d \vec l \times \vec r) }{ 4\pi { r }^{ 3 } }$ .

Since the direction is that of ( $d \vec l \times \vec r$ ), it is perpendicular to the plane containing current element and the position vector $\vec r$ .
Question 4
1 / -0

The direction of the force on a current carrying conductor held perpendicular to an uniform magnetic field is given by:

A
Fleming's right hand rule

B
Ampere's swimming rule

C
Maxwell's right hand cork screw rule

D
Fleming's left hand rule

Solution

The direction of force (motion) of a current carrying conductor in a magnetic field is given by Fleming’s left hand rule.
It states that ‘ If we hold the thumb, fore finger and middle finger of the left hand perpendicular to each other such that the fore finger points in the direction of magnetic field, the middle finger points in the direction of current, then the thumb shows the direction of force (motion) of the conductor.

Question 5

1 / -0

Match the following and find the correct pairs:

List I	List II
(a) Fleming's left hand rule	(e) Direction of induced current
(b) Right hand thumb rule	(f) Magnitude and direction of magnetic induction
(c) Biot-Savart law	(g) Direction of force due to magnetic induction
(d) Fleming's right hand rule	(h) Direction of magnetic lines due to current

a-g, b-e, c-f, d-h

a-g, b-h, c-f, d-e

a-f, b-h, c-g, d-e

a-h, b-g, c-e, d-f

Solution

a) Fleming's left hand rule is used to find the direction of force due to magnetic induction

b) Right hand thumb rule is used to find the directions of magnetic lines of force due to current

c) Biot-Savart law is used to find the magnitude and direction of magnetic induction

d) Fleming's right hand rule is used to find the direction of induced current

Question 6
1 / -0

A point charged particle of mass $2\times 10^{-4} kg$ is moving perpendicular to the uniform magnetic field of magnitude 0.1-tesla.Calculate the acceleration of the particle due to the magnetic field if its velocity is $3\times 10^4m/s$ and its charge is $+4.0\times 10^{-9}C$ .

A
0.0006 $m/s^2$

B
0.006 $m/s^2$

C
0.06 $m/s^2$

D
0.6 $m/s^2$

E
None of the above

Solution

If a be the acceleration of the particle due to only magnetic field B , then

$ma=qvB$

or $a=\dfrac{qvB}{m}$

$=\dfrac{(4\times 10^{-9})\times (3\times 10^4)\times 0.1}{2\times 10^{-4}}$

$=0.06 m/s^2$
Question 7
1 / -0

The force a magnetic field exerts on an electron is largest when the path of the electron is oriented

A
In the opposite direction from the magnetic field's direction

B
In the same direction as the magnetic field's direction

C
Up through the magnetic field at a $45^{\circ}$ angles

D
Down through the magnetic field at $45^{\circ}$ angle

E
At a right angle to the magnetic field

Solution

The magnetic force on an electron $e$ , moving with velocity $v$ in a magnetic field $B$ is given by ,
$F=evB\sin\theta$ , where $\theta$ is the angle between $v$ and $B$ ,
force $F$ will be maximum when $\sin\theta$ is maximum ,
and maximum value of $\sin\theta$ is $1$
i.e. $\sin\theta=1=\sin90$
or $\theta=90$
it means that magnetic force will be maximum when electron is moving at right angle to the magnetic field.
Question 8
1 / -0

An unknown particle is being studied in a magnetic field of variable intensity and direction. When the magnetic field is turned off, the particle is observed to move toward the earth. When the magnetic field is turned on, the particle is observed to continue to move toward the earth, no matter the strength or the direction of the magnetic field. Which of the particles listed below is most likely the unknown particle?

A
Beta particle

B
Alpha particle

C
Positron

D
Neutron

E
Gamma ray

Solution

We know that a moving charged particle experiences a force in a magnetic force , as the given unknown particle is not affected by magnetic field , so it cannot be a charged particle. But beta , alpha and positron are charged particle therefore they are not the unknown particle , only neutron is chargeless particle so it is the unknown particle.
Gamma rays are not particle though they are also not affected by a magnetic field.
Question 9
1 / -0

A cation enters in the uniform magnetic field region and moves in a semicircular path in the direction indicated in above figure. Identify the direction of the magnetic field?

A
Upward in the plane of the page

B
To the left in the plane of the page

C
To the right in the plane of the page

D
Out of the plane of the page

E
Into the plane of the page

Solution

The direction of force and velocity is shown in the figure where $F = q (v \times B)$
Using Right hand rule, the direction of magnetic field is out of the plane of paper.
Question 10
1 / -0

The diagram above shows atomic particles moving through a magnetic field.
A beam of electrons is deflected in the magnetic field and strikes at the point $P$ as shown in the figure.
At which letter would a stream of electrons strike the screen if the poles of the magnet were reversed?

A
$A$

B
$B$

C
$C$

D
$D$

E
$E$

Solution

The magnetic force on a charge $q$ in a field $\vec B$ moving with velocity $\vec v$ is given by
$\vec F=q\left(\vec v\times\vec B\right)$
if the direction of $\vec B$ is reversed than field would be $-\vec B$ due to that force will be $-\vec F$ it means the direction of force will also be reversed.Therefore now the electron beam strikes the screen at point $E$ .

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Re-Attempt Weekly Quiz Competition

Moving Charges and Magnetism Test - 51

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Moving Charges and Magnetism Test - 51

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Question 1

2r2r2r

4r4r4r

r/4r/4r/4

r/2r/2r/2

Solution

Question 2

perpendicular to the plane of the paper and outward

perpendicular to the plane of the paper and inward

towards A

towards C

Solution

Question 3

μoI(dl⃗×r⃗)4πr3\displaystyle\frac{\mu_oI (d \vec l\times \vec{r})}{4\pi r^3}4πr3μo​I(dl×r)​ perpendicular to the current element dl⃗d \vec ldl

μoI×r⃗.dl⃗4πr2\displaystyle\frac{\mu_oI\times \vec{r}.d \vec l}{4\pi r^2}4πr2μo​I×r.dl​ perpendicular to the current element dl⃗d \vec ldl

μoI×dl⃗r\displaystyle\frac{\mu_oI\times d \vec l}{ r}rμo​I×dl​ perpendicular to the plane containing the current element and position vector r⃗\vec{r}r

μoI×dl⃗4πr2\displaystyle\frac{\mu_oI\times d \vec l}{4\pi r^2}4πr2μo​I×dl​ perpendicular to the plane containing current element and position vector r⃗\vec{r}r

Solution

Question 4

Fleming's right hand rule

Ampere's swimming rule

Maxwell's right hand cork screw rule

Fleming's left hand rule

Solution

Question 5

a-g, b-e, c-f, d-h

a-g, b-h, c-f, d-e

a-f, b-h, c-g, d-e

a-h, b-g, c-e, d-f

Solution

Question 6

0.0006 m/s2m/s^2m/s2

0.006 m/s2m/s^2m/s2

0.06 m/s2m/s^2m/s2

0.6 m/s2m/s^2m/s2

None of the above

Solution

Question 7

In the opposite direction from the magnetic field's direction

In the same direction as the magnetic field's direction

Up through the magnetic field at a 45∘45^{\circ}45∘ angles

Down through the magnetic field at 45∘45^{\circ}45∘ angle

At a right angle to the magnetic field

Solution

Question 8

Beta particle

Alpha particle

Positron

Neutron

Gamma ray

Solution

Question 9

Upward in the plane of the page

To the left in the plane of the page

To the right in the plane of the page

Out of the plane of the page

Into the plane of the page

Solution

Question 10

AAA

BBB

CCC

DDD

EEE

Solution

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$2r$

$4r$

$r/4$

$r/2$

$\displaystyle\frac{\mu_oI (d \vec l\times \vec{r})}{4\pi r^3}$ perpendicular to the current element $d \vec l$

$\displaystyle\frac{\mu_oI\times \vec{r}.d \vec l}{4\pi r^2}$ perpendicular to the current element $d \vec l$

$\displaystyle\frac{\mu_oI\times d \vec l}{ r}$ perpendicular to the plane containing the current element and position vector $\vec{r}$

$\displaystyle\frac{\mu_oI\times d \vec l}{4\pi r^2}$ perpendicular to the plane containing current element and position vector $\vec{r}$

0.0006 $m/s^2$

0.006 $m/s^2$

0.06 $m/s^2$

0.6 $m/s^2$

Up through the magnetic field at a $45^{\circ}$ angles

Down through the magnetic field at $45^{\circ}$ angle

$A$

$B$

$C$

$D$

$E$