Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 66

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Question 1
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Two charge particles each of mass $$m$$ and charge $$q$$ are projected in a uniform magnetic field $$B$$ with the same speed as such planes of motion of particles are perpendicular to magnetic field $$B$$, then

A
they move on circular path of same radii

B
the magnetic force on them are same to each other

C
the kinetic energy of particles are same to each other

D
all the above

Solution

1.Magnetic force is always perpendicular to velocity therefore it does not affect the magnitude of velocity hence kinetic energy does not change.
2. As speed is same and both velocity are perpendicular to magnetic field hence magnitude of force on both charge is same.
3. Radius is given by mv/qB as all are same therefore their radius is also same.
Question 2
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Two charged particles $$M$$ and $$N$$ are projected with same velocity in a uniform magnetic field. Then $$M$$ and $$N$$ are :

A
an electron and a proton respectively

B
a deuteron and a proton respectively

C
a deuteron and a electron respectively

D
a proton and $$\alpha -$$particle respectively

Solution

Radius of charge particle in uniform magnetic filed

$$r =\dfrac{mv}{qB}$$

$$r\propto \dfrac mq$$............(1)

According to the given figure, Center of the both curve is same side, Hence nature of both side will be same.
The deuteron is neutral so it don't gate deflection.
From equation 1
$$M$$ have less radius so it will have less mass in compare to $$N$$.
Hence Option is D
Question 3
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If an electron describes half a revolution in a circle of radius $$r$$ in a magnetic field $$B$$, the energy acquire by it is :

A
$$zero$$

B
$$\dfrac {1}{2}mv^2$$

C
$$\dfrac {1}{2}\left (\dfrac {1}{2}mv^2 \right)$$

D
$$\pi r (Bev)$$

Solution

Explanation:

As per rule, the magnetic force is only responsible for changing the direction of electrons while perpendicular to velocity of electrons.
So, there would be no work done as magnetic force can not accelerate or decelerate electrons.
"Hence, energy acquired by electron will be zero"
Question 4
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The restoring couple in the moving coil galvanometer is because of :

A
magnetic field

B
material of the coil

C
twist produced in the suspension

D
current in the coil

Solution
Question 5
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A beam consisting of protons and electrons moving at the same speed goes through a thin region in which there is a magnetic field perpendicular to the beam. The protons and the electrons

A
will go undeviated

B
will be deviated by the same angle and will not separate

C
will be deviated by different angles and hence separate

D
will be deviated by the same angle but will separate

Solution

Force on a charged particle, F= qVB
For an electron, this force is F= - qVB whereas on a proton this force is F= qVB

Here, 'e' is the charge of electron. From the above formula we can see that electrons and protons will experience equal force but in opposite direction so they separate out or other words we can say that they are deviated in different directions which cause them to separate.
Question 6
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A particle of mass $$m$$ and charge $$Q$$ moving with velocity $$\vec {v}$$ describe a circles path of radius $$R$$ when subjected to a uniform transverse magnetic field of inductance $$B$$.The work done by the field when the particle completes one of circle is :

A
$$BQv(2\pi R)$$

B
$$\left (\dfrac {mv^2}{R}\right)2\pi R$$

C
$$zero$$

D
$$BQ(2\pi R)$$

Solution
Question 7
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If a charged particle of charge $$5\ \mu C$$ and mass $$5\ g$$ is moving with constant speed $$5\ m/s$$ in a uniform magnetic field $$B$$ on a curve $$x^2+y^2=25$$, where $$x$$ and $$y$$ are in metre. The value of magnetic field will be :

A
$$1$$ tesla

B
$$1$$ kilo tesla along the $$z-$$axis

C
$$5$$ kilo tesla along the $$x-$$axis

D
$$1$$ kilo tesla along any line in the $$x-y$$ plane

Solution

Along z direction
Question 8
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A charge particle follows a helical path of unequal path in a magnetic field. This means that :

A
the magnetic field non-uniform

B
the velocity vector is not parallel to the magnetic field

C
the velocity vector is not perpendicular to the magnetic field

D
all of the above

Solution
Question 9
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Directions For Questions

Ampere's law provides us an easy way to calculate the magnetic field due to a symmertrical distribution f current .its mathematical expression is $$ \oint { \overrightarrow { B} .d \overrightarrow {l} } = \mu_0 I_{in} $$
the quantity on the left hand side is known as line integral of magnetic field over a closed ampere's loop

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Only the current inside the amperian loop contributes in

A
finding magnetic field at nay point on the ampere's loop

B
line integral of magnetic field

C
in both of the above

D
in neither of them

Solution

If we are taking a current distribution field I, the ampere's loop have magnetic field due to both enclosed current distribution and outside current distribution field.
Hence, options A and C are wrong.
Now,
Magnetic field at any point on Ampere's loop can be due to all current passing through inside or outside the loop. but net contribution in the left hand side will come from inside current only. .
For r , a current passing through within the cylinder of radius r is given by
$$ \int^r_0 JdA = \int^r_0 kr^2 2 \pi rdr = 2 \pi k \int^r_0 r^3 dr $$
$$ = k \pi r^4 / 2 $$
Now using Ampere's law :
$$ B \times 2 \pi r = \mu_0 I = \mu_0 k \pi r^4 / 2 \Rightarrow B = \frac { \mu_0 k r^3 }{ 4 } $$
Hence option B is correct.
Question 10
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A charged enters a region of uniform magnetic field at an angle to the magnetic line of force . The path of the particle is a circle .

A
Circle

B
Linear

C
Helical

D
Elliptical

Solution

The velocity component $$ \nu_2$$ will be responsible in moving the charged particle in a circle .
The velocity component $$ \nu_1 $$ will be responsible in moving the charged particle in horizontal direction . Therefore , the charged particle will travel in a helical path .