Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 46

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Question 1
1 / -0

The work done by a magnetic field, on a moving charge is

A
zero because $$\vec F$$ acts parallel to $$\vec v$$

B
positive because $$\vec F$$ acts perpendicular to $$\vec v$$

C
zero because $$\vec F$$ acts perpendicular to $$\vec v$$

D
negative because $$\vec F$$ acts parallel to $$\vec v$$

Solution

Force on moving charge while moving in magnetic field is;
$$\vec = q (\vec v \times \vec B)$$
where $$\vec F$$ is perpendicular to $$\vec v$$.
Work done/sec $$= \vec F . \vec v = F \vec v cos 90^o = 0$$
Question 2
1 / -0

Two infinite long wires, each carrying current $$I$$, are lying along x-axis and y-axis, respectively. $$A$$ charged particle, having a charge $$q$$ and mass $$m$$, is projected with a velocity $$u$$ along the straight line $$OP$$. The path of the particle is (neglect gravity) a:

A
straight line

B
circle

C
helix

D
cycloid

Solution

Magnetic field due to the vertical wire is equal and opposite to the magnetic field due to the horizontal wire at any point on line OP, hence, they will cancel each other.

Hence, the charged particle projected along OP will not experience any force due to $$B=0$$, it will move along the straight line OP.
Question 3
1 / -0

A positively charged particle moving due east enters a region of uniform magnetic field directed vertically upwards. This particle will

A
get deflected in vertically upward direction

B
move in circular path with an increased speed

C
move in a circular path with decreased speed

D
move in a circular path with uniform speed

Solution

When a positively charged enters in a region of uniform magnetic field directed vertically upwards, it experiences a centripetal force which moves the particle in circular path with a uniform speed (in clockwise direction).
Question 4
1 / -0

If an electron describes half a revolution in a circle of radius r in a magnetic field B, the energy acquired by it is

A
zero

B
$$\displaystyle \frac{1}{2} mv^2$$

C
$$\displaystyle \frac{1}{4} mv^2$$

D
$$\mu_r \times Bev$$

Solution

Since magnetic force is always perpendicular to the velocity of electron.
So it can only change the direction of velocity of electron.
But it (the magnetic force) cannot accelerate or deaccelerate the electron.
Question 5
1 / -0

Through two parallel wires A and B, 10A and 2A of currents are passed respectively in opposite directions. If the wire A is infinitely long and the length of the wire B is 2m, then force on the conductor B, which is situated at 10 cm distance from A, will be

A
$$8 \times 10^{-7} N$$

B
$$8 \times 10^{-5} N$$

C
$$4 \times 10^{-7} N$$

D
$$4 \times 10^{-5} N$$

Solution

$$\displaystyle F = \dfrac{\mu_0}{4 \pi} \dfrac{2 I_1 I_2}{r} \times l $$

$$= \dfrac{10^{-7} \times 2 \times 10 \times 2}{0.1} \times 2 $$

$$= 8 \times 10^{-5} N$$
Question 6
1 / -0

Directions For Questions

Ampere's law provides us an easy way to calculate the magnetic field due to symmetrical distribution of current. Its mathematical expression is $$\oint { \vec { B } .d\vec { l } } ={ \mu }_{ 0 }{ I }_{ in }$$
The quantity on the left hand side is known as line integral of magnetic field over a closed Ampere's loop.

...view full instructions

Only the current inside the Amperian loop contributes in

A
finding magnetic field at any point on the Ampere's loop

B
line integral of magnetic field

C
in both of the above

D
in neither of them

Solution

Magnetic field at any point on Ampere's lop can be due to all currents passing through inside or outside the loop. But net contribution in the left hand side will come form inside current only.
For $$r<a$$ current passing through within the cylinder of radius $$r$$ is given by

$$\int _{ 0 }^{ r }{ JdA } =\int _{ 0 }^{ r }{ k{ r }^{ 2 }2\pi rdr } =2\pi k\int _{ 0 }^{ r }{ { r }^{ 3 }dr } =k\pi { r }^{ 4 }/2$$

Now using Ampere's law:

$$B\times 2\pi r={ \mu }_{ 0 }I={ \mu }_{ 0 }k\pi { r }^{ 4 }/2\quad \Rightarrow B=\cfrac { { \mu }_{ 0 }k{ r }^{ 3 } }{ 4 } $$
Question 7
1 / -0

Two thin, long, parallel wires, separated by a distance 'd' carry a current of 'i' A in the same direction. They will

A
repel each other with a force of $$\mu_0 i^2/(2 \pi d)$$

B
attract each other with a force of $$\mu_0 i^2 / (2\pi d)$$

C
repel each other with a force of $$\mu_0i^2 / (2 \pi rd^2)$$

D
attract each other with a force of $$\mu_0 i^2 / (2 \pi d^2)$$

Solution

$$\displaystyle \dfrac{F}{l} = \dfrac{\mu_0 i_1 i_2}{2 \pi d} $$

$$= \dfrac{\mu_0 i^2}{2 \pi d}$$

(attractive as current is in the same direction)
Question 8
1 / -0

Two thin long parallel wires separated by a distance $$b$$ are carrying a current $$I$$ ampere each. The magnitude of the force per unit length exerted by one wire on the other is

A
$$\dfrac {\mu_0I^2}{b^2}$$

B
$$\dfrac {\mu_0I^2}{2\pi b}$$

C
$$\dfrac {\mu_0I}{2\pi b}$$

D
$$\dfrac {\mu_0I}{2\pi b^2}$$

Solution

Given, $$i_1=i_2=i$$

$$ \therefore F=\dfrac{\mu_0i^2l}{2\pi b}$$

Hence, force per unit length is $$F=\dfrac{\mu_0i^2}{2\pi b}$$
Question 9
1 / -0

There wires are situated at the same distance. A current of $$1\ A$$, $$2\ A$$, $$3\ A$$ flows through these wires in the same direction. What is ratio $$F_1/ F_2$$ where $$F_1$$ is force on $$1$$ and $$F_2$$ on $$2$$?

A
$$7/8$$

B
$$1$$

C
$$9/8$$

D
none of the above

Solution

Force per unit length is given as $$F = \dfrac{\mu_\circ{}\times I_1\times I_2}{2 \pi d}$$

Force on wire 1 due to wires 2 and 3 is given as

$$F_1 = I_1 \times (\dfrac{\mu_\circ{}I_2}{2 \pi d}$$ $$ +\dfrac{\mu_\circ{}I_3}{2 \pi 2d})$$

Given $$I_1 =1 A, I_2 =2 A$$, $$I_3 =3 A$$

$$F_1 = \dfrac{\mu_\circ{}2}{2 \pi d}$$ $$ +\dfrac{\mu_\circ{}3}{4 \pi d}$$

$$F_1 = \dfrac{7\mu_\circ{}}{4 \pi d}$$

$$I_1 =1 A$$, $$I_3 =3 A$$
Force on wire 2 due to wires 1 and 3 is given as

$$F_3 = I_2 A\times (\dfrac{\mu_\circ{}I_1}{2 \pi d}$$ $$ -\dfrac{\mu_\circ{}I_3}{2 \pi d})$$

$$F_2 = 2 A\times (\dfrac{\mu_\circ{}1}{2 \pi d}$$ $$ -\dfrac{\mu_\circ{}3}{2 \pi d})$$

$$F_2 = -\dfrac{4\mu_\circ{}}{2 \pi d}$$

$$\dfrac{F_1}{F_2}=\dfrac{7}{8}$$
Question 10
1 / -0

An electron travelling with a speed u along the positive x-axis enters into a region of magnetic field where $$B=-B_0 \hat k(x > 0)$$. It comes out of the region with speed v. Then

A
$$v=u$$ at y > 0

B
$$v=u$$ at y < 0

C
$$v > u$$ at y > 0

D
$$v > u$$ at y < 0

Solution

$$\vec{F} =q(\vec{v} \times \vec{B})=-e(u\hat{i} \times B(-\hat{k}))=euB(-\hat{j})$$
The spped will not change in a magnetic field.
The electron will not have $$y \gt 0$$ during its course of motion.

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Re-Attempt Weekly Quiz Competition

Moving Charges and Magnetism Test - 46

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Moving Charges and Magnetism Test - 46

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Question 1

zero because $$\vec F$$ acts parallel to $$\vec v$$

positive because $$\vec F$$ acts perpendicular to $$\vec v$$

zero because $$\vec F$$ acts perpendicular to $$\vec v$$

negative because $$\vec F$$ acts parallel to $$\vec v$$

Solution

Question 2

straight line

circle

helix

cycloid

Solution

Question 3

get deflected in vertically upward direction

move in circular path with an increased speed

move in a circular path with decreased speed

move in a circular path with uniform speed

Solution

Question 4

zero

$$\displaystyle \frac{1}{2} mv^2$$

$$\displaystyle \frac{1}{4} mv^2$$

$$\mu_r \times Bev$$

Solution

Question 5

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

$$8 \times 10^{-5} N$$

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

$$4 \times 10^{-5} N$$

Solution

Question 6

finding magnetic field at any point on the Ampere's loop

line integral of magnetic field

in both of the above

in neither of them

Solution

Question 7

repel each other with a force of $$\mu_0 i^2/(2 \pi d)$$

attract each other with a force of $$\mu_0 i^2 / (2\pi d)$$

repel each other with a force of $$\mu_0i^2 / (2 \pi rd^2)$$

attract each other with a force of $$\mu_0 i^2 / (2 \pi d^2)$$

Solution

Question 8

$$\dfrac {\mu_0I^2}{b^2}$$

$$\dfrac {\mu_0I^2}{2\pi b}$$

$$\dfrac {\mu_0I}{2\pi b}$$

$$\dfrac {\mu_0I}{2\pi b^2}$$

Solution

Question 9

$$7/8$$

$$1$$

$$9/8$$

none of the above

Solution

Question 10

$$v=u$$ at y > 0

$$v=u$$ at y < 0

$$v > u$$ at y > 0

$$v > u$$ at y < 0

Solution

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