Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 59

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Weekly Quiz Competition

Question 1
1 / -0

The incorrect statement of the following is:

A
Work done on a moving charge in a magnetic field is zero

B
The kinetic energy of a moving charge in magnetic field is conserved

C
Magnetic field is not produced due to electron moving in a straight line

D
Kinetic energy of a moving charge in an electric field changes

Solution

Movement of charge in a straight line results in a magnetic field being generated from the relation $$F=qvB\sin\theta$$
Question 2
1 / -0

A proton and an alpha particle are separately projected in a region , where a uniform magnetic field exists. The initial velocities are perpendicular to the direction of magnetic field . If both the particles move along circles of equal radii , the ratio of momentum of proton to alpha particle( $$\dfrac{P_p}{P_a}$$)

A
1

B
1/2

C
2

D
1/4

Solution
Question 3
1 / -0

If an electron of velocity $$(2\hat{i}+3\hat{j})$$ is subjected to magnetic field of $$4\hat{k}$$, then its:-
(a) path will change
(b) speed does not change
(c) path must be circular
(d) momentum is constant.

A
a, b

B
All

C
a, b, c

D
None

Solution

As shown, the Lorentz Force on the electron is: $$e(\vec{V} \times \vec {B})$$
In this case, it is given by:
$$e((2\hat { i } +3\hat { j } )\times 4\hat { k } )=e(3\hat { i } -8\hat { j } )$$

which shows that the force acts at some angle to the velocity.
Hence its path would change.

Also, a magnetic field always acts perpendicular to s=displacement, therefore no work is done. Hence, the change in kinetic energy is zero (Work-Energy theorem). Therefore the speed will not change.
Question 4
1 / -0

A charged particle moves in a uniform magnetic field perpendicular to it, with a radius of curvature 4 cm. On passing through a metallic sheet it looses half of its kinetic energy, then the radius of curvature of the particle is :

A
2 cm

B
4 cm

C
8 cm

D
$$2\sqrt2$$ cm

Solution

$$R= \dfrac{mv}{qB} =\dfrac{m}{qB}\sqrt{\dfrac{2(KE)}{m}}= \dfrac{\sqrt{2m}}{qB}\sqrt{KE}$$

$$\Rightarrow \dfrac{R_1}{R_2}= \dfrac{\sqrt{KE}}{\sqrt{KE/2}}$$

$$\Rightarrow R_2= \dfrac{R_1}{\sqrt2}=\dfrac{4}{\sqrt2}= 2\sqrt2cm$$
Question 5
1 / -0

The coil of a tangent galvanometer is put in the magnetic meridian

A
To avoid the magnetic effect of the earth's field

B
To produce intense magnetic field at the centre of the coil

C
To produce a field at right angle to the earth's field

D
To avoide error due to parallax

Solution
Question 6
1 / -0

The charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field $$B={ B }_{ 0 }\hat { k }$$.

A
They have equal z-components of momenta

B
They must have equal charges

C
They necessarily represent a particle, antiparticle pair

D
The charge to mass ratio satisfy

$$\left( \dfrac{q_1}{m}\right)_{1}+\left( \dfrac{q_2}{m}\right)_{2}=0$$

Solution

As the two charged particles travers identical helical path the pitchof their helices will be same
Now pitch will be same
Now pitch $$=$$ Time for one circle completion $$\times$$ speed
$$ \Rightarrow P = T \times v\cos \theta = \dfrac{2\pi m}{qB} v\cos\theta$$ (Time period under uniform magnetic field $$= \dfrac{2\pi m}{qB})$$
As $$ P_{1}=P_{2}$$
$$ \Rightarrow \dfrac{2\pi m_{1}}{q_{1}B_{0}} v\cos \theta = \dfrac{2\pi m_{2}}{q_{2}B_{0}} v\cos\theta $$
$$ \Rightarrow \dfrac{q_{1}}{m_{1}} = \dfrac{q_{2}}{m_{2}}$$
So the have equal charge to mass eatio magnitude
But they are moving in opposite direction so one is $$+Ve$$ charge and another $$-Ve$$
$$ \Rightarrow \dfrac{q_{1}}{m_{1}} + \dfrac{q_{2}}{m_{2}} = 0 $$
Question 7
1 / -0

A positively charged particle moving with constant speed due east enters a region of uniform magnetic field directed vertically upwards this particle will-

A
Get deflected vertically upwards direction

B
move in a circular path with its increased speed

C
move in a circular path with its unchanged speed

D
Be moving in a straight part with the same speed

Solution

By right-hand rule force, $$\vec F= q\vec V \times \vec B$$ will be towards south so the charge will start moving in a circular path.
As the force is in the perpendicular direction so it will not do any work so speed will not change.
Option C is correct.
Question 8
1 / -0

A charged particle of mass 1 $$\mu $$ g and charge 1 $$\mu $$ C having speed 2m/s moves in a magnetic field 1 $$\hat{K} $$ Tesla. At a certain instant its velocity is $$\bar{V}=\hat{i}+\hat{j}+m\hat{k} $$ . The value of 'm' is ________________.

A
(1) $$\sqrt{2}$$

B
(2) 0

C
(3) 1

D
(4) Can't determine
Question 9
1 / -0

A current carrying wire and a rectangular loop are placed as shown in figure then-

A
The wire will be repelled by the rectangular loop

B
The wire will attracted by the rectangular loop

C
There will not be any change in position of wire

D
Loop will rotate

Solution

Since, there is an attractive force between $$PS$$ and the adjacent wire and the repulsive force between $$QR$$ and the adjacent wire but $$B\alpha \dfrac1{R}\implies$$ the attractive force prevails and rectangular loop is attracted to the adjacent current wire.
Question 10
1 / -0

The magnetic field strength at a point P distant r due to an infinite straight wire as shown in the figure carrying i is :

A
$$\mu_0$$

B
$$\mu_0i/ 2\sqrt2r$$

C
$$(\mu_0i/ \sqrt2 \pi r)$$

D
$$\frac{\mu_oi}{4\pi r}[2+\sqrt2]$$

Solution

We know,

Magnetic field due to semi-infinite wire,
$$B=\dfrac{\mu_0I}{4\pi R}(\cos\theta_1+\cos\theta_2)$$

$$B_1=\dfrac{\mu_0I}{4\pi R}(1+\cos(45^0))$$

$$B_2=\dfrac{\mu_0I}{4\pi R}(1+\cos(45^0))$$

$$\overrightarrow B_{net}=\overrightarrow {B_1}+ \overrightarrow{B_2}$$

$$\overrightarrow B=\dfrac{\mu_0 I}{4\pi R}(2+\sqrt 2)$$

Option $$\textbf D$$ is the correct answer.

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Moving Charges and Magnetism Test - 59

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Question 1

Work done on a moving charge in a magnetic field is zero

The kinetic energy of a moving charge in magnetic field is conserved

Magnetic field is not produced due to electron moving in a straight line

Kinetic energy of a moving charge in an electric field changes

Solution

Question 2

1

1/2

2

1/4

Solution

Question 3

a, b

All

a, b, c

None

Solution

Question 4

2 cm

4 cm

8 cm

$$2\sqrt2$$ cm

Solution

Question 5

To avoid the magnetic effect of the earth's field

To produce intense magnetic field at the centre of the coil

To produce a field at right angle to the earth's field

To avoide error due to parallax

Solution

Question 6

They have equal z-components of momenta

They must have equal charges

They necessarily represent a particle, antiparticle pair

The charge to mass ratio satisfy$$\left( \dfrac{q_1}{m}\right)_{1}+\left( \dfrac{q_2}{m}\right)_{2}=0$$

Solution

Question 7

Get deflected vertically upwards direction

move in a circular path with its increased speed

move in a circular path with its unchanged speed

Be moving in a straight part with the same speed

Solution

Question 8

(1) $$\sqrt{2}$$

(2) 0

(3) 1

(4) Can't determine

Question 9

The wire will be repelled by the rectangular loop

The wire will attracted by the rectangular loop

There will not be any change in position of wire

Loop will rotate

Solution

Question 10

$$\mu_0$$

$$\mu_0i/ 2\sqrt2r$$

$$(\mu_0i/ \sqrt2 \pi r)$$

$$\frac{\mu_oi}{4\pi r}[2+\sqrt2]$$

Solution

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The charge to mass ratio satisfy

$$\left( \dfrac{q_1}{m}\right)_{1}+\left( \dfrac{q_2}{m}\right)_{2}=0$$