Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 45

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Question 1
1 / -0

The magnetic filed (dB) due to smaller element (dl) at a distance $$(\vec r)$$ from element carrying current i, is

A
$$\displaystyle dB = \frac{\mu_0 i}{4 \pi} \left ( \frac{\vec{dl} \times \vec r}{r} \right )$$

B
$$\displaystyle dB = \frac{\mu_0 i}{4 \pi} i^2 \left ( \frac{\vec{dl} \times \vec r}{r^2} \right )$$

C
$$\displaystyle dB = \frac{\mu_0 i}{4 \pi} i^3 \left ( \frac{\vec{dl} \times \vec r}{2r^2} \right )$$

D
$$\displaystyle dB = \frac{\mu_0}{4 \pi} i \left ( \frac{\vec{dl} \times \vec r}{r^3} \right )$$

Solution

$$dB=\dfrac { { \mu }_{ 0 }i }{ 4\pi } \int { \dfrac { \left( \overrightarrow { dl } \times \hat { r } \right) }{ { r }^{ 2 } } } \\$$

we know that=$$\hat { r } =\dfrac { \overrightarrow { r } }{ { r }} \\$$

$$dB=\dfrac { { \mu }_{ 0 }i }{ { 4 }\pi } \int { \dfrac { \left( \overrightarrow { dl } \times \overrightarrow { r } \right) }{ { r }^{ 3 } } }$$
Question 2
1 / -0

Three long, straight and parallel wires are arranged as shown in figure. The force experienced by $$10cm$$ length of wire $$Q$$ is

A
$$1.4\times {10}^{-4}N$$ toward the right

B
$$1.4\times {10}^{-4}N$$ toward the left

C
$$2.6\times {10}^{-4}N$$ toward the right

D
$$2.6\times {10}^{-4}N$$ toward the left

Solution

Magnetic field produced by wire R at Q: $$B_1 = \dfrac{\mu_0i_1}{2\pi r_1} = \dfrac{\mu_0 \times 20}{2\pi \times ( 2 \times 10^{-2})}$$

Magnetic field produced by wire P at Q: $$B_2 = \dfrac{\mu_0i_2}{2\pi

r_2} = \dfrac{\mu_0 \times 30}{2\pi \times ( 10 \times 10^{-2})}$$

$$B_{net} =B_1 -B_2= \dfrac{\mu_0 \times 7}{2\pi \times 10^{-2}}$$

Force on 10 cm of wire at Q: $$F_Q =i_Ql B_{net}$$

$$F=\cfrac { 4\pi \times { 10 }^{ -7 }\times 20\times 10\times 10\times { 10 }^{ -2 } }{ 2\pi \times 10\times { 10 }^{ -2 } } $$

$$=\cfrac { 4\pi \times { 10 }^{ -7 }\times 10\times { 10 }^{ -2 } }{ 2\pi \times { 10 }^{ -2 } } \left[ 100-30 \right] $$

$$=20\times { 10 }^{ -7 }\times 70=1400\times { 10 }^{ -7 }$$

$$=1.4\times { 10 }^{ -4 }N$$ toward right.
Question 3
1 / -0

A current $$i$$ ampere flows along an infinitely long straight thin walled tube, then the magnetic induction at any point inside the tube is

A
$$\infty$$

B
$$zero$$

C
$$\displaystyle \dfrac{\mu_0}{4 \pi}. \frac{2i}{r}\ tesla$$

D
$$\displaystyle \dfrac{2i}{r}\ tesla$$

Solution

Magnetic induction inside a thin walled tube is zero.
(According to Ampere's law)
Question 4
1 / -0

A charge particle enters into a magnetic field with a velocity vector making an angle of $$30^o$$ with respect to the direction of magnetic field. The path of the particle is:

A
circular

B
helical

C
elliptical

D
straight line

Solution

Here velocity vector have two components
(i) v cos $$\theta$$, parallel to magnetic field
(ii) v sin $$\theta$$, perpendicular to magnetic field. Due to component v cos $$\theta$$, the particle will have a linear motion but due to v sin $$\theta$$, the particle will have simultaneously a circular motion. The resultant of the two is a helical path.
Question 5
1 / -0

An electron is ejected from the surface of a long, thick straight conductor carrying a current, initially in a direction perpendicular to the conductor. The electron will

A
ultimately return to the conductor

B
move in a circular path around the conductor

C
gradually move away from the conductor along a spiral

D
move in a helical path, with the conductor as the axis

Solution

The initial force on the electron is downward.
As the electron changes direction, the force on it remains in the xy-plane, with a component directed toward the conductor
Question 6
1 / -0

A current of $$\dfrac{1}{(4\pi)}$$ ampere is flowing in a long straight conductor. The line integral of magnetic induction around a closed path enclosing the current-carrying conductor is

A
$${10}^{-7}Wb$$ $${m}^{-1}$$

B
$$4\pi \times{10}^{-7}Wb$$ $${m}^{-1}$$

C
$$16{\pi}^{2} {10}^{-7}Wb$$ $${m}^{-1}$$

D
$$zero$$

Solution

$$\oint { \vec { B } } \cdot d\vec { l } ={ \mu }_{ 0 }I$$

$$=4\pi \times {

10 }^{ -7 }\times \cfrac { 1 }{ 4\pi } $$

$$= { 10 }^{ -7 }Wb{ m }^{

-1 }$$
Question 7
1 / -0

A stream of electrons is projected horizontally to the right. A straight conductor carrying a current is supported parallel to the electron stream and above it. If the current in the conductor is from left to right, what will be the effect on the electron stream?

A
The electron stream will be pulled upward

B
The electron stream will be pulled downward

C
The electron stream will be retarded

D
The electron stream will be speeded up toward the right

Solution

Current due to a stream of electrons will be in a direction opposite to the direction of current in the straight conductor which is held parallel above the stream of electrons. The two currents are in opposite directions and hence they will repel.
So, the stream of electrons is pulled downward.
Question 8
1 / -0

If long hollow copper pipe carries a direct current, the magnetic field associated with the current will be:

A
only inside the pipe

B
only outside the pipe

C
neither inside nor outside the pipe

D
both inside and outside the pipe

Solution

According to Ampere's law $$\int \vec B \bullet \vec d l = \mu_0i$$
To any closed path inside pipe, there will be no current enclosed.
So B inside pipe is zero.
Only, there would be field outside the pipe.
Question 9
1 / -0

Two parallel wires carrying currents in the same direction attract each other because of

A
potential difference between them

B
mutual inductance between them

C
electric forces between them

D
magnetic forces between them

Solution

Let wire on RHS carry current vertically upwards and wire on LHS carries current vertically upwards then the direction of magnetic field at a position where RHS wire is there due to LHS wire is inside the plane. Now RHS wire is a current carrying conductor placed in a magnetic field due to LHS wire. Similarly LHS wire is a current carrying conductor placed in a magnetic field due to RHS wire. Therefore, they both will experience force.
Now in RHS wire current is vertically upwards and direction of magnetic field is inside the plane. Since direction of force is given by $$F=\overrightarrow {dl}\times \overrightarrow B$$ therefore, force acting on RHS wire is towards left and by applying similar method we find that force acting on LHS wire is towards right. Therefore, they both attract each other and the cause is magnet forces acting between them.
Question 10
1 / -0

A small current element of length $$dl$$ and carrying current is placed at $$(1, 1, 0)$$ and is carrying current in '$$+z$$' direction. If magnetic field at origin be $$\overrightarrow{B}_1$$ and at point $$(2, 2, 0)$$ be $$\vec B_2$$ then

A
$$\overrightarrow{B} = \overrightarrow{B}$$

B
$$|\overrightarrow{B}_1 | = | 2\overrightarrow{B}_2 |$$

C
$$\overrightarrow{B}_1 = - \overrightarrow{B}_2$$

D
$$\overrightarrow{B}_1 = - 2\overrightarrow{B}_2$$

Solution

Now it is given that a small current carrying element is placed at $$(1,1,0)$$ and carrying current in $$+z$$ direction.
Then by using right hand thumb rule we can find the magnetic field lines which would be circular in this case having center at $$(1,1,0)$$
Now since origin and $$(2,2,0)$$ are two points which will lie on a circular centered at $$(1,1,0)$$ therefore, magnitude of magnetic field will be same at both the points.
And we find that origin and $$(2,2,0)$$ are diametrical opposite points therefore, direction of magnetic field at origin will be just opposite of the direction of magnetic field at $$(2,2,0)$$

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Moving Charges and Magnetism Test - 45

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Question 1

$$\displaystyle dB = \frac{\mu_0 i}{4 \pi} \left ( \frac{\vec{dl} \times \vec r}{r} \right )$$

$$\displaystyle dB = \frac{\mu_0 i}{4 \pi} i^2 \left ( \frac{\vec{dl} \times \vec r}{r^2} \right )$$

$$\displaystyle dB = \frac{\mu_0 i}{4 \pi} i^3 \left ( \frac{\vec{dl} \times \vec r}{2r^2} \right )$$

$$\displaystyle dB = \frac{\mu_0}{4 \pi} i \left ( \frac{\vec{dl} \times \vec r}{r^3} \right )$$

Solution

Question 2

$$1.4\times {10}^{-4}N$$ toward the right

$$1.4\times {10}^{-4}N$$ toward the left

$$2.6\times {10}^{-4}N$$ toward the right

$$2.6\times {10}^{-4}N$$ toward the left

Solution

Question 3

$$\infty$$

$$zero$$

$$\displaystyle \dfrac{\mu_0}{4 \pi}. \frac{2i}{r}\ tesla$$

$$\displaystyle \dfrac{2i}{r}\ tesla$$

Solution

Question 4

circular

helical

elliptical

straight line

Solution

Question 5

ultimately return to the conductor

move in a circular path around the conductor

gradually move away from the conductor along a spiral

move in a helical path, with the conductor as the axis

Solution

Question 6

$${10}^{-7}Wb$$ $${m}^{-1}$$

$$4\pi \times{10}^{-7}Wb$$ $${m}^{-1}$$

$$16{\pi}^{2} {10}^{-7}Wb$$ $${m}^{-1}$$

$$zero$$

Solution

Question 7

The electron stream will be pulled upward

The electron stream will be pulled downward

The electron stream will be retarded

The electron stream will be speeded up toward the right

Solution

Question 8

only inside the pipe

only outside the pipe

neither inside nor outside the pipe

both inside and outside the pipe

Solution

Question 9

potential difference between them

mutual inductance between them

electric forces between them

magnetic forces between them

Solution

Question 10

$$\overrightarrow{B} = \overrightarrow{B}$$

$$|\overrightarrow{B}_1 | = | 2\overrightarrow{B}_2 |$$

$$\overrightarrow{B}_1 = - \overrightarrow{B}_2$$

$$\overrightarrow{B}_1 = - 2\overrightarrow{B}_2$$

Solution

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