Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 52

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Question 1
1 / -0

Which line shows the path of alpha radiation in a magnetic field?

A
A

B
B

C
C

D
D

E
E

Solution

If magnetic field is directed ,perpendicularly inward the plane of paper then being positive , alpha particle would experience a force towards left to the line $$A$$ (according to Fleming's left hand rule) and as it is heavier than beta particle therefore it will be deflected less due to magnetic force and go through the line $$B$$ , and if the direction of magnetic field is perpendicularly outward the plane of paper , alpha particle will go through the line $$C$$.
Question 2
1 / -0

Two wires are aligned side by side and both hooked into different circuits in which the current is allowed to flow into each circuit. Identify which of the following statements is correct?

A
The two currents destructively interfere with one another

B
The two currents constructively interfere with one another

C
The two wires attract and move closer

D
The two wires repel and move away

E
The two wires remain still

Solution

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Question 3
1 / -0

A proton enters a region of constant magnetic field, $$B$$, of magnitude $$1.0$$ tesla with initial speed of $$1.5 \times {10}^{6} {m}/{s}$$ . If the protons initial velocity vector makes an angle of $${30}^{o}$$ with the direction of $$B$$. Calculate the speed of the protons $$4$$ seconds after entering the magnetic field.

A
$$5.0 \times {10}^{5} {m}/{s}$$

B
$$7.5 \times {10}^{5} {m}/{s}$$

C
$$1.5 \times {10}^{6} {m}/{s}$$

D
$$3.0 \times {10}^{6} {m}/{s}$$

E
$$6.0 \times {10}^{6} {m}/{s}$$

Solution

The magnetic force does not change the speed of a charged particle, it only changes the velocity i.e. the direction of charged particle ,
The magnetic force on a charge $$q$$ is given by $$\vec{F}=q\left(\vec{v}\times \vec{B}\right)$$
The acceleration by this force on of mass m of proton will be$$\vec{a}=q\left(\vec{v}\times\vec{B}\right)/m$$
The speed will change only if any component of $$\vec{a}$$ is in the direction of velocity but cross product shows that acceleration is perpendicular to velocity so can only change the direction of velocity and speed will remain same as $$1.5\times{10}^{6} m/s$$
Question 4
1 / -0

Which line shows the path of beta radiation in a magnetic field?

A
A

B
B

C
C

D
D

E
E

Solution

If magnetic field is directed ,perpendicularly inward the plane of paper then being negative , beta particle would experience a force towards right to the line $$A$$ (according to Fleming's left hand rule) and as it is lighter than alpha particle therefore it will be deflected more due to magnetic force and go through the line $$E$$ , and if the direction of magnetic field is perpendicularly outward the plane of paper , alpha particle will go through the line $$D$$.
Question 5
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 protons strike the screen?

A
$$A$$

B
$$B$$

C
$$C$$

D
$$D$$

E
$$E$$

Solution

A proton has the charge opposite to that of electron so it will go in opposite direction of electron at point $$D$$ besides that direction of force on a charge $$q$$ can be determined by Fleming's left hand rule or by
$$\vec F=q\left(\vec v\times \vec B\right)$$
if $$\vec v$$ is in $$y $$ direction ,$$\vec B $$ is in $$x$$ direction then $$\vec F$$ will be in $$z$$ direction
Further a proton is heavier than electron therefore its deflection will be smaller as compared to electron and proton strikes to point $$D$$ instead of point $$E$$.
Question 6
1 / -0

A positively charged particle of charge $$q$$ is moving in circular path of radius $$r$$ within uniform magnetic field, $$B$$. Find the expression which gives the magnitude of the particle's linear momentum.

A
$$qBr$$

B
$$\dfrac{qB}{r}$$

C
$$\dfrac{q}{(Br)}$$

D
$$\dfrac{B}{(qr)}$$

E
$$\dfrac{r}{(qB)}$$

Solution

Here the centripetal force will balance the magnetic force.

So, $$mv^2/r=qvB$$

or $$mv=qBr$$

Thus, linear momentum, $$p=mv=qBr$$
Question 7
1 / -0

A positively charged particle of $$1.0 C$$ is moving upward at a velocity of $$2.0 \times {10}^{3} {m}/{s}$$ in a magnetic field of strength $$4.0 \times {10}^{-4} T$$, directed into the page. Calculate the magnetic force experienced by the charge particle.

A
$$0.8 N$$ to the left

B
$$0.8 N$$ to the right

C
$$2.0 \times {10}^{-7} N$$ to the left

D
$$2.0 \times {10}^{-7} N$$ to the right

E
$$5.0 \times {10}^{6} N$$ to the left

Solution

Given : $$B = 4.0 \times 10^{-4} T$$ $$Q = 1.0 C$$ $$v = 2.0 \times 10^3$$ $$m/s$$
Magnetic force $$\vec{F} = Q (\vec{v} \times \vec{B})$$
Thus from figure, $$(\vec{v} \times \vec{B})$$ points in the left direction, thus force acts to the left.
Force on the particle $$|F| = QvB = 1\times 2\times 10^3 \times 4 \times 10^{-4} = 0.8 N$$ (to the left)
Question 8
1 / -0

In the above shown figure, what is the direction of the magnetic force $$F_B$$ ?

A
To the right

B
Downward, in the plane of the page

C
Upward, in the plane of the page

D
Out of the plane of the page

E
Into the plane of the page

Solution

Magnetic force $$\vec{F_B} =-q (\vec{v} \times \vec{B}) = q (\vec{B} \times \vec{v})$$
Using right hand thumb rule, we get that $$(\vec{B} \times \vec{v})$$ points in the direction out of the plane of page.
Hence the direction of magnetic force is out of the plane of page.
Question 9
1 / -0

Two electrons are entering perpendicularly to the magnetic field. The velocity of one of the electrons is three times greater than the velocity of the other electron. Calculate the ratio of the circular radii of the path followed by the two electrons ?

A
The faster electron has a radius two times larger than the slower electron

B
The faster electron has a radius three times larger than the slower electron

C
The faster electron has a radius eight times larger than the slower electron

D
The faster electron has a radius sixteen times larger than the slower electron

E
The faster electron has a radius sixty-four times larger than the slower electron

Solution

Using $$Bqr = mv$$
$$\implies r = \dfrac{mv}{Bq}$$
$$\implies$$ $$r \propto v$$
Given : $$v' = 3v$$
$$\therefore$$ $$\dfrac{r'}{r} =\dfrac{v'}{v} =3$$
$$\implies r' =3r$$
Thus the fast electron has a radius three times larger than the slower electron.
Question 10
1 / -0

A point charge particle of charge q and mass m is moving in circular path of radius r with constant speed v in a uniform magnetic field due to magnetic force. If the speed is doubled to 2 v then what happens to the period of revolution, T?

A
T increases by a factor of 2

B
T increases by a factor of 4

C
T stays the same

D
T decreases by a factor of 2

E
T decreases by a factor of 4

Solution

The force due to magnetic field on the charged particle provides the centripetal acceleration to the charge.

Thus $$qvB=\dfrac{mv^2}{r}$$

$$\implies \dfrac{r}{v}=\dfrac{m}{qB}$$

The time period of revolution is $$T=\dfrac{2\pi r}{v}$$

$$=\dfrac{2\pi m}{qB}$$

which is independent of the speed of the particle.
Thus when the velocity becomes twice, the radius also becomes twice to keep the time period of revolution the same.