Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 74

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Weekly Quiz Competition

Question 1
1 / -0

Positive point chages $$q=+8.00\mu C$$ and $$q'=+3.00\mu C$$ are moving relative to an observer at point $$P$$, as shown in the figure. the distance $$d$$ is $$0.120$$ $$m$$. When the two charges are at the locations as shown in the figure, what are the magnitude and direction of the net magnetic field they produce at point $$P$$?
(Take $$v=4.50\times { 10 }^{ 6 }{ m{ s }^{ -1 }}$$ and $$v'=9.00\times { 10 }^{ 6 }{ m{ s }^{ -1 }}$$).

A
$$4.38\times { 10 }^{ -4 }$$ $$T$$, into the page

B
$$4.38\times { 10 }^{ -4 }$$ $$T$$, out of the page

C
$$2.16\times { 10 }^{ -4 }$$ $$T$$, into the page

D
$$2.9\times { 10 }^{ -9 }$$ $$T$$, out of the page

Solution

$${ B }_{ total }=B+B'=\cfrac { { \mu }_{ 0 } }{ 4\pi } \left( \cfrac { qv }{ { d }^{ 2 } } +\cfrac { q'v' }{ { d }^{ 2 } } \right) $$
$$B=\cfrac { { \mu }_{ 0 } }{ 4\pi } \left( \cfrac { (8.0\times { 10 }^{ -6 }C)(4.5\times { 10 }^{ 6 }{ ms }^{ -1 }) }{ { \left( 0.120m \right) }^{ 2 } } +\cfrac { (3.0\times { 10 }^{ -6 }C)(9.0\times { 10 }^{ 6 }{ ms }^{ -1 }) }{ { \left( 0.120m \right) }^{ 2 } } \right) $$
$$B=4.38\times { 10 }^{ -4 }T$$
Direction is into the page.
Question 2
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Figure shows an Amperian path $$ABCDA$$. Part $$ABC$$ is in verical plane $$PSTU$$ while part $$CDA$$ is in horizontal plane $$PQRS$$. Direction of circulation along the path is shown by an arrow near point $$B$$ and $$D$$.
$$\oint { \vec { B } .d\vec { l } } $$ for this path according to Ampere's law will be :

A
$$\left( { I }_{ 1 }-{ I }_{ 2 }+{ I }_{ 3 } \right) { \mu }_{ 0 }$$

B
$$\left( -{ I }_{ 1 }+{ I }_{ 2 } \right) { \mu }_{ 0 }$$

C
$${ I }_{ 3 }{ \mu }_{ 0 }$$

D
$$\left( { I }_{ 1 }+{ I }_{ 2 } \right) { \mu }_{ 0 }$$

Solution

$$\mu_0(I_{enc})=\oint \bar B.\bar dl$$
Area vector of ABC is along the right direction, while ACD is along the upward direction.
Now currents passing through ABC are $$I_1,I_3$$ along the area directions.
Now through ACD, $$I_2,I_3; I_2$$ along the area vector while $$I_3$$ is opposite to it.
Thus, $$\mu_0(I_1+I_3+I_2-I_3) = \oint \bar B.\bar dl = \mu_0 (I_1+I_2)$$
Question 3
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A very long straight conducting wire lying along the z-axis carries a current of $$2A$$. The integral $$\oint { \vec { B } .d\vec { l } } $$ is computed along the straight line $$PQ$$, where $$P$$ has the coordinates $$(2cm,0,0)$$ and $$Q$$ has the coordintes $$(2cm,2cm,0)$$. The integral has the magnitude (in SI units).

A
$$\cfrac {\pi}{2}\times { 10 }^{ -7 }$$

B
$${8\pi}\times { 10 }^{ -7 }$$

C
$${2\pi}\times { 10 }^{ -7 }$$

D
$${\pi}\times { 10 }^{ -7 }$$
Question 4
1 / -0

A conducting bar with mass m and length L slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current I in the rails and bar, and a constant, uniform, vertical magnetic field $$\vec B$$ fills the region between the rails (as shown in fig). Find the magnitude and direction of the net force on the conducting bar. Ignore friction, air resistance and electrical resistance :

A
ILB, to the right

B
ILB, to the left

C
2ILB, to the right

D
2ILB, to the left

Solution

The force acting on the current carrying rod placed in a magnetic field is given by
$$\vec{F}=I(\vec{l}\times \vec{B})$$
$$=ILB$$ towards right
Question 5
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A magnetic field $$\vec B=B_0\hat j$$ exists in the region $$a < x < 2a$$ and $$\vec B=-B_0\hat j$$, in the region 2a < x < 3a, where $$V_0$$ is a positive constant. A positive point charge moving with a velocity $$\vec v=v_0\hat i$$, where $$v_0$$ is a positive constant, enters the magnetic field at $$x=a$$. The trajectory of the charge in this region can be like

A

B

C

D

Solution

Force on a charged particle $$\vec{F} = q( \vec{v} \times \vec{B})$$
From a to 2a:
$$\vec{B} = B_0\hat{j}$$, $$\vec{v}=v_0\hat{i}$$
$$ \therefore \vec{F} = qv_0B_0(\hat{i} \times \hat{j})= qv_0B_0\hat{k}$$
Thus the particle gets deflected towards z-axis.
From 2a to 3a:
$$\vec{B} = -B_0\hat{j}$$, $$\vec{v}=v_0\hat{i}$$
$$ \therefore \vec{F} = -qv_0B_0(\hat{i} \times \hat{j})= -qv_0B_0\hat{k}$$
Thus the particle gets deflected towards -ve z-axis.
Since the particle enters at $$x=a$$, the particle will have z-coordinate as zero.
Question 6
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Two long parallel wires are separated by a distance of $$2.50 cm$$.The force per unit length that each wire exerts on the other is $$4.00\times {10}^{-5}N/m$$, and the wires repel each other. The current in one wire is $$0.600 A$$. What is the current in the second wire?

A
$$8.33A$$

B
$$9.56A$$

C
$$5.45A$$

D
$$6.88A$$

Solution

$$\displaystyle\frac{F}{l}=(\frac{{\mu}_{o}}{2\pi})\frac{{i}_{1}{i}_{2}}{r}$$

$$ 4\times {10}^{-5}=\displaystyle\frac{(2\times {10}^{-7})(0.6){i}_{2}}{2.5\times {10}^{-2}}$$

$$\therefore {i}_{2}=8.33 A$$
Question 7
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Consider the three long,straight,parallel wires as shown in figure.Find the force experienced by a 25 cm length of wire C.

A
$$2\times 10^{-4}\ N$$

B
$$3\times 10^{-4}\ N$$

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

D
$$6\times 10^{-4}\ N$$

Solution

Repulsion by wire D ...... (towards right)

$${F}_{1}=\displaystyle\frac{{\mu}_{o}}{2\pi}\frac{{i}_{1}{i}_{2}l}{r}$$

$$=\displaystyle\frac{(2\times {10}^{-7})(30\times 10)}{3\times {10}^{-2}}(0.25)$$

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

Repulsion by wire G ....... (towards left)

$${F}_{2}=\displaystyle\frac{(2\times {10}^{-7})(20\times 10)}{5\times {10}^{-2}}(0.25)$$

$$ =2\times {10}^{-4}N$$

$$\therefore {F}_{net}={F}_{1} -{F}_{2}$$

$$=3\times {10}^{-4}N$$ (towards right)
Question 8
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Three wires $$A,B$$, and $$C$$ all are all parallel and have the same magnitude of current. Wires $$A$$ and $$C$$ have current going toward the top of the screen while wire $$B$$ (which is in the middle) has current going toward the bottom of the screen.
What is the direction of the force on $$A$$ because of the other two wires?

A
To the right.

B
To the left.

C
Into the screen.

D
Out of the screen.

Solution

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Question 9
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$$AB$$ is a long wire carrying a current $${I}_{1}$$ and $$PQRS$$ is a rectangular loop carrying currect $${I}_{2}$$ (as shown in the figure)
which among the following statements are correct?
(a) Arm $$PQ$$ will get attracted to wire $$AB$$, and the arm $$RS$$ will get repelled from wire $$AB$$
(b) Arm $$PQ$$ will get repelled from wire $$AB$$ and arm $$RS$$ attracted to wire $$AB$$
(c) Forces on the arm $$PQ$$ and $$RS$$ will be unequal and opposite
(d) Forces on the arm $$PQ$$ and $$RS$$ will be zero

A
Only (a)

B
(b) and (d)

C
(a) and (c)

D
(b) and (c)

Solution

$$\because$$ current in wire $$AB$$ and $$PQ$$ are in the same direction. Therefore they will attract
$$\because$$ current in wire $$AB$$ and $$RS$$ are in opposite direction. Therefore they will repeal each other
magnetic field around $$PQ$$ more than $$RS$$. So force on the arms $$PQ$$ and $$RS$$ will be unequal and opposite
Question 10
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An electric current carrying conductor produces .......... field around it

A
Magnetic

B
Electric

C
Gravitational

D
All of these