Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 64

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Weekly Quiz Competition

Question 1
1 / -0

A conductivity rod of mass 50 gm and length 10 cm slide without friction on two long. horizontal roads. A uniform magnetic field of magnitude 5 mT exists in the region as shown. A source S is used to maintain a counter current 2 A through the rod. If motion of the rod starts from the rest, its speed after 10s from the start of the motion will be

A
1 cm/s

B
8 cm/s

C
12 cm/s

D
20 cm/s

Solution

$$\textbf{Solution}:$$
The rod will move towards left with acceleration
a=$$\dfrac{ilB}{m}$$
Therefore velocity is given by,
$$v=(\dfrac{ilB}{m})t$$= ($$\dfrac{2\times10\times5}{50})10$$
= 20 cm/s

$$\textbf{Hence D is the correct option}$$
Question 2
1 / -0

A charge particle of mass m and charge q is projected with velocity v along plane at a distance a from a long staright current carrying conductor as shown in figure. the radius of curvature of the path traced by the particle at the given position does not depend on

A
$$ \theta $$

B
q

C
i

D
m

Solution

$$Given:$$ mass=$$m$$ ; charge=$$q$$ ; velocity =$$v$$ ; distance=$$a$$
$$Solution:$$ radius of curvature
$$r=\dfrac{mV} {qB}$$
So, $$B = \dfrac {\mu _o i} {2 \pi a}$$

$$r = \dfrac{mV 2 \pi a} {q \mu _ o i}$$
so, independent of $$\theta$$
$$So,the $$ $$correct$$ $$option:A$$
Question 3
1 / -0

A particle of charge $$1\mu C$$ is at rest in a magnetic field $$\vec{B}= -2\hat{k}$$ tesla. Magnetic Lorentz force on the charge particle with respect to an observer moving with velocity $$\vec{v}= -5\hat{i} ms^{-1}$$ will be

A
Zero

B
$$-10^{-5}\hat{j} N$$

C
$$-10^{-6}\hat{j} N$$

D
$$+10^{-5}\hat{j} N$$

Solution
Question 4
1 / -0

A straight current carrying conductor is kept along the axis of circular loop carrying current. The force exerted by the straight conductor on the loop is:

A
Perpendicular to the plane of the loop

B
Zero

C
In the plane of the loop, away from the centre

D
In the plane of the loop, towards the centre

Solution

Given
A straight current carrying wire is placed on the axis of circular loop carrying current
Solution
Force of one conductor on other is iLBsin$$\theta$$
Here Magnetic field of straight wire is in same direction as of length of circular loop.
Therefore$$sin\theta=sin0=0$$
F=0
The correct option is B
Question 5
1 / -0

Electric field and magnetic field in a region of space are given by $$ \overrightarrow E = E_0 \hat j $$ and $$ \overrightarrow B = B_0 \hat j $$ . A charge particle is released at origin with velocity $$ \overrightarrow v =v_0 \hat k $$ then path of particle is

A
Straight line

B
Helical with uniform pitch

C
Circular path

D
Helical with increasing pitch

Solution

$$Given:$$ Electric field =$$\overrightarrow{ E }= E _{ o } \hat{ j }$$ ; magnetic field = $$\overrightarrow{ B }= B _{ o } \hat{ j }$$ ; velocity= $$\overrightarrow{ v }= v _{ o } \hat{ j }$$
$$Solution:$$ Electric field (acting along $$\hat{ j }$$ direction) will change the velocity component which is parallel to $$\overrightarrow{ B }$$ (which is also along $$\overrightarrow{ j }$$ direction). $$\overrightarrow{ B }= B _{ o } \hat{ j }$$ and $$\overrightarrow{ v }= v _{ o } \hat{ j }$$ will rotate the particle in
a circle.
Hence the net path is helical with increasing pitch.
$$So,the$$ $$correct$$ $$option:D$$
Question 6
1 / -0

A charged particle moves through a magnetic field in a direction perpendicular to it. Then the:

A
acceleration remains unchanged

B
B. velocity remains unchanged

C
speed of the particle remains unchanged

D
direction of the particle remains unchanged

Solution

Solution: When a charged particle move inside the magnetic field in the direction perpendicular to the direction of motion, it moves in a circular motion. While moving in a circular motion, at every point the direction of the particle changes, so its velocity keeps changing at every point (But, the magnitude remains constant).
So, the speed of the particle remains unchanged.
Hence, the correct option is (C).
Question 7
1 / -0

A straight wire carrying current 10A is lying such that position vector at its ends are $$\vec r_1 = 2 \hat i - 2 \hat j$$ and $$\vec y_2 = 10 \hat i + 4 \hat j$$. It is kept in a uniform magnetic field 1T $$\hat k$$. Then the force acting on wire is :-

A
$$50 N$$

B
$$100 N$$

C
$$150 N$$

D
$$200 N$$

Solution

Given: $$\overrightarrow r_1=2i-2j$$
  $$\overrightarrow r_2=10i+4j$$
  $$\overrightarrow B=1k$$
here, $$i,j,k$$ are all unit vectors.
$$I=10A$$
To find: force acting on wire,$$||\overrightarrow F||=?$$
Solution: As we know that,
$$\overrightarrow F=I.(\overrightarrow l*\overrightarrow B)$$ $$...(1)$$
Now, $$\overrightarrow l=\overrightarrow r_2-\overrightarrow r_1$$
   $$==>\overrightarrow l=10i+4j-2i+2j$$
$$==>\overrightarrow l=8i+6j$$
substitute required values in $$eq^n(1)$$,we get
$$==>\overrightarrow F=10.(8i+6j)*(1k)$$
$$==>\overrightarrow F=10.(6i-8j)$$
Now, magnitude of F is,
$$==>||\overrightarrow F||=\sqrt{10^2*(6^2+8^2)}$$
$$==>||\overrightarrow F||=\sqrt{100*(36+64)}$$
$$==>||\overrightarrow F||=\sqrt{100*100}$$
$$==>||\overrightarrow F||=100N$$
hence,
The correct opt : B
Question 8
1 / -0

Two infinitely long parallel wires carry equal current in same direction. The magnetic field at a mid point in between the two wires is

A
Twice the magnetic field produced due to each of the wires

B
Half of the magnetic field produced due to each of the wires

C
Square of the magnetic field produced due to each of the wires

D
Zero

Solution

At midpoint, magnetic fields due to both the wires are equal and opposite. So $$B=0$$.
Question 9
1 / -0

A charge $$'q'$$ moves in a region where electric field $$'E'$$ nd magnetic filed $$'B'$$ both exist, then force on it is :-

A
$$q(\vec{V}\times \vec{B})$$

B
$$q\vec{E}$$

C
$$q\left\{\vec{E}+(\vec{v}\times \vec{B}) \right\} $$

D
$$q\left\{\vec{B}+(\vec{v}\times \vec{E}) \right\} $$

Solution

Solution: While moving in a space, the force experienced by the charged particle in the region where magnetic field and electric field exists is called Lorentz Force.
The force experienced by a charge particle of charge "q" moving in an electric field is given by,
$$\overrightarrow{F_{e}}=q\overrightarrow{E}$$
The direction of force depends on the nature of charge. If positive, the direction of force is in the direction of electric field, and if negative the direction of force is opposite to the direction of electric field.
The force experienced by a charge particle moving in the magnetic field $$\overrightarrow{B}$$ with velocity $$\overrightarrow{v}$$ is given by,
$$\overrightarrow{F_{m}}=q(\overrightarrow{v}\times \overrightarrow{B})$$
The direction of force is in the direction of $$(\overrightarrow{v}\times \overrightarrow{B})$$ given by right hand screw rule.
So, the total force experienced by charge particle is given by,
$$\overrightarrow{F}= \overrightarrow{F_{e}}+\overrightarrow{F_{m}}=q(\overrightarrow{E}+(\overrightarrow{v}\times \overrightarrow{B}))$$
Hence, the correct option is (C).
Question 10
1 / -0

A proton, an electron, and a Helium nucleus, have the same energy. They are in circular orbits in a plane due to magnetic field perpendicular to the plane. let $$r_p$$, $$r_e$$ and $$r_{He}$$ be their respectively radii, then ,

A
$$r_e > r_p > r_{He}$$

B
$$r_e < r_p < r_{He}$$

C
$$r_e < r_p = r_{He}$$

D
$$r_e > r_p = r_{He}$$

Solution

$$r=\dfrac{mv}{qB}=\dfrac{\sqrt{2mK}}{qB}$$
$$r_{He}=r_p > r_e$$

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Moving Charges and Magnetism Test - 64

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Moving Charges and Magnetism Test - 64

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Question 1

1 cm/s

8 cm/s

12 cm/s

20 cm/s

Solution

Question 2

$$ \theta $$

q

i

m

Solution

Question 3

Zero

$$-10^{-5}\hat{j} N$$

$$-10^{-6}\hat{j} N$$

$$+10^{-5}\hat{j} N$$

Solution

Question 4

Perpendicular to the plane of the loop

Zero

In the plane of the loop, away from the centre

In the plane of the loop, towards the centre

Solution

Question 5

Straight line

Helical with uniform pitch

Circular path

Helical with increasing pitch

Solution

Question 6

acceleration remains unchanged

B. velocity remains unchanged

speed of the particle remains unchanged

direction of the particle remains unchanged

Solution

Question 7

$$50 N$$

$$100 N$$

$$150 N$$

$$200 N$$

Solution

Question 8

Twice the magnetic field produced due to each of the wires

Half of the magnetic field produced due to each of the wires

Square of the magnetic field produced due to each of the wires

Zero

Solution

Question 9

$$q(\vec{V}\times \vec{B})$$

$$q\vec{E}$$

$$q\left\{\vec{E}+(\vec{v}\times \vec{B}) \right\} $$

$$q\left\{\vec{B}+(\vec{v}\times \vec{E}) \right\} $$

Solution

Question 10

$$r_e > r_p > r_{He}$$

$$r_e < r_p < r_{He}$$

$$r_e < r_p = r_{He}$$

$$r_e > r_p = r_{He}$$

Solution

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