Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 48

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Weekly Quiz Competition

Question 1
1 / -0

The acceleration of an electron at a certain moment in a magnetic field $$\vec{B}=2\hat{i}+3\hat{j}+4\hat{k}$$ is $$\vec{a}=x\hat{i}+\hat{j}-\hat{k}$$.The value of x is

A
0.5

B
1

C
2.5

D
1.5

Solution

$$\vec{B}\cdot \vec{F}=0$$ as $$\vec{B}\bot \vec{F}$$
$$\therefore \vec{B}\cdot \vec{\alpha} =0$$

$$ 2x+3-4=0$$
$$\therefore x=0.5$$
Question 2
1 / -0

A charged particle moving along positive x-direction with a velocity $$v$$ enters a region where there is a uniform magnetic field $$\vec{B}=-B\hat{k}$$,from x=0 to x=d.The particle gets deflected at an angle $$\theta$$ from its initial path.The specific charge of the particle is

A
$$\displaystyle\frac{Bd}{vcos\theta}$$

B
$$\displaystyle\frac{vtan\theta}{Bd}$$

C
$$\displaystyle\frac{Bsin\theta}{vd}$$

D
$$\displaystyle\frac{vsin\theta}{Bd}$$

Solution

$$sin\theta =\displaystyle\dfrac{d}{r}$$

$$=\dfrac{d}{(mv/Bq)}$$

$$\therefore \displaystyle\dfrac{q}{m}=\dfrac{vsin\theta}{Bd}$$
Question 3
1 / -0

If a charged particle projected from origin with some initial velocity $$({v}_{0}\hat { i } )$$ in a uniform magneitc field $$({B}_{0}\hat { i } )$$. At some point of time direction of magnetic field is reversed then which of the following path is/are possible for the charged particle. Here $${v}_{0}$$ and $${B}_{0}$$ are positive constants.

A

B

C

D
None of the above

Solution

Here as we know that the force due to magnetic field on a particle
is $$ F = q (v\times B) $$
so the force of magnetic field is zero if the velocity of the particle is in the same direction as of the magnetic field so as in this
here $$ v = v_{o} i $$ and $$ B = B_{o} i $$
so no force on the particle even if you reverse the direction of the magnetic field.
so particle will always move with a constant velocity in a straight line
so none of the option is correct.
Question 4
1 / -0

The force of repulsion between two parallel wires is $$F$$ when each one of them carries a certain current I. If the current in each is doubled, the force between them would be

A
$$8F$$

B
$$4F$$

C
$$2F$$

D
$$F$$

Solution

The force per unit length on a current carrying wire due to another current carrying wire placed at a distance $$d$$ from it is

$$\dfrac{F}{l}=\dfrac{\mu_0I_1I_2}{2\pi d}$$

Hence when current in both wires is doubled, the force of repulsion becomes four times of that before.
Question 5
1 / -0

Two long parallel wires are at a distance of 1m. If both of them carry one ampere of current in same direction, then the force of attraction on unit length of the wires will be

A
$$2\times 10^{-7}N/m$$

B
$$4\times 10^{-7}N/m$$

C
$$8\times 10^{-7}N/m$$

D
$$10^{-7}N/m$$

Solution

Given: $$I_1= I_2 = 1$$ A
Distance between the wires, $$d = 1$$ m

Force of attraction per unit length, $$f = \dfrac{\mu_0 I_1 I_2}{2\pi d}$$ $$N/m$$

$$\therefore $$ $$f = \dfrac{4\pi \times 10^{-7} \times (1) (1)}{2 \pi \times 1} = 2 \times 10^{-7}$$ $$N/m$$
Question 6
1 / -0

A rectangular loop carrying a current $$i_1$$, is situated near a long straight wire carrying a steady current $$i_2$$. The wire is parallel to one of the sides of the loop and is in the plane of the loop as shown in the figure. Then the current loop will

A
move away from the wire

B
move towards the wire

C
remain stationary

D
rotate about an axis parallel to the wire

Solution

The magnetic field due to the straight current current carrying wire decreases with increasing distance from the wire. Thus the magnetic field is stronger at the side closer to wire and weaker at the other.
Force on each segment of wire is $$BIL$$
Hence force acting on nearer side towards the wire is greater than the force acting on farther side away from wire. Hence the net force on loop is towards the wire.
Question 7
1 / -0

When a charged particle particle, moving with a velocity v, is subjected to a magnetic field, the force on it is non-zero. This implies that

A
Angle between them is either $$0^{\circ}$$ and $$180^{\circ}$$

B
Angle between them is necessarily $$90^{\circ}$$

C
Angle between them can have any value other than $$90^{\circ}$$

D
Angle between them can have any value other than $$0^{\circ}$$ and $$180^{\circ}$$

Solution

The force acting on a charge particle moving with velocity $$\vec{v}$$ in a magnetic field $$\vec{B}$$ is given by

$$\vec{F}=q(\vec{v}\times \vec{B})$$

$$\implies F=qvBsin\theta$$

which is non-zero for $$0^{\circ}\neq\theta\neq180^{\circ}$$
Question 8
1 / -0

A magnetic field cannot exert any force on a :

A
Moving magnet

B
Stationary magnet

C
Moving charge

D
Stationary charge

Solution

Hint: Use formula of Lorentz force
Explanation:
Force on charge q with velocity $$\vec{v}$$ by magnetic field $$\vec{v}$$ is given by the formula:
$$\vec{F} = q\ (\vec{v}\ X\ \vec{B})$$
As given that not any force is exerted, so cross product of $$\vec{v}$$ and $$\vec{B}$$ should be zero. Which is possible if both are parallel to each other or value of one of those is zero. Value of magnetic field is not zero here so velocity must be zero.
And for only stationary particles velocity is zero.
Answer:
Hence, option D is the correct answer.
Question 9
1 / -0

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

A
speed of the particle remains unchanged

B
direction of the particle remains unchanged

C
acceleration remains unchanged

D
velocity remains unchanged

Solution

$$\vec{F}=q\vec{v}\times\vec{B}$$
When $$\vec{v}$$ is perpendicular to $$\vec{B}$$, the charged particle moves in a circular path in a plane perpendicular to $$\vec{B}$$ and the magnetic force $$\vec{F}$$ always act towards the center of the circle with magintude,
$$F=qvB$$
Thus, the direction of $$|vec{v}$$ changes but not the magnitude.
Hence, speed of particle remains unchanged.
Question 10
1 / -0

When a charged particle moving with velocity $$\vec V$$ is subjected to a magnetic field of induction $$\vec B$$, the force on it is non-zero. This implies the

A
Angle between $$\vec V$$ and $$\vec B$$ is necessary $$90^o$$

B
Angle between $$\vec V$$ and $$\vec B$$ can have a value other than $$90^o$$

C
Angle between $$\vec V$$ and $$\vec B$$ can have a value other than zero and $$180^o$$

D
Angle between $$\vec V$$ and $$\vec B$$ is either zero or $$180^o$$

Solution

Force on a charged particle moving in a magnetic field is

$$\vec{F}=q(\vec{v}\times \vec{B})$$

Let the angle between $$\vec{v}$$ and $$\vec{B}$$ be $$\theta$$

Hence $$\left|\vec{F}\right|=qvBsin\theta \neq 0$$

$$\implies sin\theta\neq 0$$

$$\implies \theta\neq 0^{\circ},180^{\circ}$$

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Moving Charges and Magnetism Test - 48

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Moving Charges and Magnetism Test - 48

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Question 1

0.5

1

2.5

1.5

Solution

Question 2

$$\displaystyle\frac{Bd}{vcos\theta}$$

$$\displaystyle\frac{vtan\theta}{Bd}$$

$$\displaystyle\frac{Bsin\theta}{vd}$$

$$\displaystyle\frac{vsin\theta}{Bd}$$

Solution

Question 3

None of the above

Solution

Question 4

$$8F$$

$$4F$$

$$2F$$

$$F$$

Solution

Question 5

$$2\times 10^{-7}N/m$$

$$4\times 10^{-7}N/m$$

$$8\times 10^{-7}N/m$$

$$10^{-7}N/m$$

Solution

Question 6

move away from the wire

move towards the wire

remain stationary

rotate about an axis parallel to the wire

Solution

Question 7

Angle between them is either $$0^{\circ}$$ and $$180^{\circ}$$

Angle between them is necessarily $$90^{\circ}$$

Angle between them can have any value other than $$90^{\circ}$$

Angle between them can have any value other than $$0^{\circ}$$ and $$180^{\circ}$$

Solution

Question 8

Moving magnet

Stationary magnet

Moving charge

Stationary charge

Solution

Question 9

speed of the particle remains unchanged

direction of the particle remains unchanged

acceleration remains unchanged

velocity remains unchanged

Solution

Question 10

Angle between $$\vec V$$ and $$\vec B$$ is necessary $$90^o$$

Angle between $$\vec V$$ and $$\vec B$$ can have a value other than $$90^o$$

Angle between $$\vec V$$ and $$\vec B$$ can have a value other than zero and $$180^o$$

Angle between $$\vec V$$ and $$\vec B$$ is either zero or $$180^o$$

Solution

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