Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 63

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Question 1
1 / -0

Two parallel wires carry currents of 20 A and 40 A in opposite directions. Another wire carrying a current antiparallel to 20 A is placed midway between the two wires. The magnetic force on it will be

A
towards 20 A

B
towards 40 A

C
zero

D
perpendicular to the plane of the currents

Solution
Question 2
1 / -0

A conducting rod $$MN$$ of mass $$'m$$ and length $$'\ell '$$ is placed on parallel smooth conducting rails connected to an uncharged capacitor of capacitance $$'C'$$ and a battery of emf $$\varepsilon $$ as shown. A uniform magnetic field $$'B'$$ is existing perpendicular to the plane of the rails.The steady state velocity acquired by the conducting rod $$MN$$ after closing switch $$S$$ is (neglect the resistance of the parallel rails and the conducting rod )

A
$$\frac{{2CB\ell \varepsilon }}{{\left( {m + C{B^2}{\ell ^2}} \right)}}$$

B
$$\frac{{CB\ell \varepsilon }}{{\left( {m + C{B^2}{\ell ^2}} \right)}}$$

C
$$\frac{{CB\ell \varepsilon }}{{2\left( {m + C{B^2}{\ell ^2}} \right)}}$$

D
$$\frac{{CB\ell \varepsilon }}{{4\left( {m + C{B^2}{\ell ^2}} \right)}}$$

Solution

Given :Mass=m ,Length =l, Capacitance=C, Emf=$$\varepsilon $$,Magnetic field =B
Solution: Let the potential difference across conducting rod=$$V_{1}$$
Let the potential difference across capacitor=$$V_{2}$$
So,$$\varepsilon=V_{1}+V_{2}$$
$$\varepsilon=Bvl+\frac{q}{C} $$
$$q=\left ( \varepsilon -Bvl \right )C $$......(1)
Force required to pull the loop with constant velocity:
$$ F=ma=m\frac{dv}{dt}$$
$$ Bil=m\frac{dv}{dt}$$
Integrating the euation:
$$m\int_{0}^{v}dv=Bl\int_{0}^{q}dq\times dt $$
$$ m\left ( v-0 \right )=Blq$$
$$q=\frac{mv}{Bl}$$.....(2)
From equation (1) & (2):
$$\left ( \varepsilon -Bvl \right )C=\frac{mv}{Bl} $$
The steady state velocity acquired by the conducting rod MN after closing switch S is:
$$v=\frac{C\varepsilon Bl}{m+B^{2}l^{2}C^{2}}$$
Hence the correct option is B
Question 3
1 / -0

A magnetic field is applied at perpendicular to the axis of cylinderical discharge tube then cathode and positive rays deflected in

A
same direction

B
opposite direction

C
Mutually perpendicular direction

D
No deflection for both

Solution

$$\textbf{Solution}:$$
1. The rays in a cathode ray tube are streams of electrons.
2.Electrons are negatively charged, and a magnetic field exerts forces on electrically charged particles that are in motion in a direction other than that of the magnetic field.
3. When a magnetic field is applied over the discharge tube, the cathode rays gets deflected towards the positively charged plate. This deviation is because of the fact that cathode rays are negatively charged.

$$\textbf{Hence A is the correct option}$$
Question 4
1 / -0

If two streams of electrons move parallel to each other in same direction then they.

A
attract each other

B
repel each other

C
do not exert any force on one other

D
Get rotated perpendicular to each other

Solution

For change particle, if they are moving freely in space, electrostatic force is dominant over magnetic force between them. Hence due to electric force they repel each other.
Question 5
1 / -0

Two infinitely long parallel wires carrying current of 3A and 5A placed in the same plane at a distance of 2 m. The magnetic forces on 5 cm segment of wire A is:

A
$$1.5 \times{10}^{-8}N$$

B
$$7.5 \times{10}^{-8}N$$

C
$$3 \times{10}^{-5}N$$

D
zero

Solution

$$\textbf{Given}:$$ $$I_{1}=3A, I_{2}=5A, l = 0.05 m, r = 2m$$

$$\textbf{Solution}:$$
We know that the force between two infintely long wire is given by,
$$ F=\dfrac{\mu_{o}}{4\pi}\times\dfrac{2I_{1}I_{2}}{r}\times l $$
$$=10^{−7}\times\dfrac{2\times5\times3\times0⋅05}{(2⋅0)}$$
=$$7.5\times10^{-8}N $$

$$\textbf{Hence B is the correct option}$$
Question 6
1 / -0

Two parallel wire carrying current in opposite direction

A
attract each other

B
repel each other

C
no force act between them

D
none of them

Solution
Question 7
1 / -0

One gram hydrogen has $$6\times 10^{23}$$ atoms. Imagine that all the nuclei are put at the north pole of the earth and the electrons at the south pole of the earth (radius=6400 km). The force between the charges is

A
$$10\times 10^{5}N$$

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

C
$$2.5\times 10^{5}N$$

D
$$2\times 10^{5}N$$

Solution

Given: Mass of hydrogen= 1 gram ;Radius=6400 km
To Find: Forces between charges
Solution : No of mole=$$\frac{weight}{molecular weight}=\frac{1}{2}$$
No of electrons and protons=$$2*\frac{1}{2}*6*10^{23}$$
=$$6*10^{23}$$
Charge(q)=ne=$$1.6*10^{-19}*6*10^{23}$$
=960000
Force=$$\frac{kq}{r^{2}}^{2}$$
=$$\frac{9*10^{^{9}}*\left \{ \right.9.6*10^{4}\left. \right \}^{2}}{6400^{2}}$$
=$$10*10^{5}$$N
Answer: Hence option A is correct
Question 8
1 / -0

Two parallel wire carries current $$I_1$$ and $$I_2$$ are separated by distance $$d$$. Force per unit length of wire is $$F$$.
Then :

A
$$F\propto d$$

B
$$f\propto \dfrac{1}{d}$$

C
$$F\propto d^2$$

D
$$F\propto \dfrac{1}{d^2}$$

Solution

Given : Two parallel wire carries current I1 and I2 are separated by distance d. Force per unit length of wire is F.

Solution:
We have the formula for force between two wires carrying current $$I_{1} and I_{2} $$

$$\frac{dF}{dL}=\frac{\mu I_{1}I_{2}}{2\pi d}$$
From here we can conclude that $$f\propto \frac{1}{d}$$
The Correct Opt=B
Question 9
1 / -0

A charged particle enters a magnetic field at right angles to the magnetic filed as shown in figure. the field exits for a length equal to $$1.25$$ times the radius of circular path of the particle. the deviation of the particle is

A
$$ 90^0 $$

B
$$ 180^0 $$

C
$$ 270^0 $$

D
$$ 300^0 $$
Question 10
1 / -0

The charge on a particle is $$100$$ times that of electron. It is revolving in a circular path of radius $$0.8 \mathrm { m }$$ at a frequency of $$10 ^ { th }$$ revolutions per second. The magnetic field at the centre of path will be -

A
$${10^{-16}} \mu_0 $$

B
$$\dfrac {10^{-7}}{ \mu} $$

C

$${10^{-17}} \mu $$

D
$${10^{-6}} \mu $$

Solution

$$\begin{array}{l} T=0.1\, \, \sec \\ i=\dfrac { { 100\times e } }{ { 0.1 } } =1000e \\ B=\dfrac { { \mu oi } }{ { 2r } } =\dfrac { { { \mu _{ 0 } }\times 1000\times 1.6\times { { 10 }^{ -19 } } } }{ { 2\times 0.8 } } \\ ={ 10^{ -16 } }{ \mu _{ 0 } } \end{array}$$