Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 60

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Weekly Quiz Competition

Question 1
1 / -0

A proton enters a magnetic field of flux density 1.5 T velocity of $$20 \times 10^2 ms ^{-1}$$ at an angle of $$30 ^\circ$$ with the field. The force on the position is $$[e_p = 1.6 \times 10^{-19} C]$$

A
$$2.4 \times 10^{-10} N$$

B
$$2.403 \times 10^{-12} N$$

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

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

Solution

The magnitude of the magnetic force on a charge particle $$F = \left| q \right|\,v\,B\sin \theta $$
Here, q=charge on a proton $$ = 1.602 \times {10^{ - 19}}C$$
$$\begin{array}{l} V=2\times { 10^{ 7 } }\, m/s \\ \theta ={ 30^{ 0 } } \\ B=1.5\, T \end{array}$$
Substituting all these values in the expression for $$F$$
We get,
$$\begin{array}{l} F=1.602\times { 10^{ -19 } }C\times 2\times { 10^{ 7 } }\, m/s\, \times 1.5\, T\sin { 30^{ 0 } } \\ =2.403\times { 10^{ -12 } }N \end{array}$$
$$\therefore$$ Option $$B$$ is correct.
Question 2
1 / -0

When a charged particle enters in a uniform magnetic field, then its kinetic energy

A
remains constant

B
increases

C
decreases

D
becomes zero

Solution

When a charged particle enters a magnetic field $$B$$ its kinetic energy remains constant as the force exerted on the particle is:
$$F=q\overrightarrow { V } \times \overrightarrow { B } $$
is perpendicular to $$\overrightarrow {V}$$, so the work done by $$\overrightarrow {B}=0$$. This does not cause any change in kinetic energy.
Question 3
1 / -0

Two parallel wires on of length $$1\ m$$ and other is infinitive, are lying at a distance of $$2m$$. If the current flowing in each wire is $$1$$ ampere then the force between them will be-

A
$$2 \times 10^{-7}\ N$$

B
$$10^{-7}\ N$$

C
$$0.5\ N$$

D
$$10^{7}\ N$$

Solution

The force between the two parallel wires one of length $$1m$$ and other in infinite,
$$F=\dfrac{\mu_0I_1I_2l}{2\pi d}$$. . . . . .(1)
Given,
$$I_1=I_2=1A$$
$$d=2m$$
$$l=1m$$
$$\mu_0=4\pi\times 10^{-7}$$
Put all value in equation (1), we get
$$F=\dfrac{4\pi\times 10^{-7}\times 1\times 1 \times 1}{2\pi \times 2}$$
$$F=10^{-7}N$$
The correct option is B.
Question 4
1 / -0

A particle of mass 0.5 gm and charge $$2.5\times 10^{-8} C$$ is moving with velocity $$6\times 10^4 m/s$$. What should be the minimum value of magnetic field acting on it, so that the particle is able to move in a straight line ? $$ (g=9.8 m/sec^2)$$

A
$$0.327 Weber/m^2$$

B
$$3.27 Weber/m^2$$

C
$$32.7 Weber/m^2$$

D
none of these

Solution

We know force= mass$$\times$$acceleration due to gravity=0.5$$\times$$ 9.8 m/$${s}^{2}$$.
Thus force=4.9 N
Also we know Force$$F=q[v\times B]$$
$$B=\dfrac{4.9}{2.5\times 6\times {10}^{4}}=0.327 $$
Question 5
1 / -0

The maximum energy of a deuteron coming out a cyclotron is $$20\ MeV$$. The maximum energy of proton that can be obtained from this accelerator is:

A
$$10\ MeV$$

B
$$20\ MeV$$

C
$$30\ MeV$$

D
$$40\ MeV$$

Solution

In cyclotron output energy of particles coming out for deutron is

$$ {{E}_{d}}=\dfrac{{{q}^{2}}{{B}^{2}}{{R}^{2}}}{2{{m}_{D}}} $$

$$ {{E}_{P}}=\dfrac{{{q}^{2}}{{B}^{2}}{{R}^{2}}}{2{{m}_{P}}} $$
Taking ratios,

$$ \dfrac{{{E}_{d}}}{{{E}_{p}}}=\dfrac{{{m}_{p}}}{{{m}_{d}}} $$ $$ \because {{m}_{p}}=1.6\times {{10}^{-27}}\,\,\And \,\,{{m}_{d}}=3.2\times {{10}^{-27}} $$

$$ \dfrac{20}{{{E}_{P}}}=\dfrac{1.6\times {{10}^{-27}}}{3.2\times {{10}^{-27}}} $$

$$ {{E}_{p}}=40\,\,MeV $$
Question 6
1 / -0

Which of the following particles will have minimum frequency of revolution when projected with the same velocity perpendicular to a magnetic field?

A
electron

B
proton

C
$$He^+$$

D
$$Li^+$$

Solution
Question 7
1 / -0

Two conducting coils are placed co-axially now a cell is connected in one coil then they will :-

A
Attract to each other

B
Repel to each other

C
Both (1) & (2)

D
They will not experience any force

Solution

Due to Lenz law, the other coil will 'try' to oppose this flux by a current passing through the other coil which will decrease the magnetic flux because of the first coil.
Question 8
1 / -0

A particle of mass m, charge $$q$$ and kinetic energy $$t$$ enters a transverse uniform magnetic field of induction B. After 3 s, the kinetic energy of the particle will be

A
$$3T$$

B
$$2T$$

C
$$T$$

D
$$4T$$

Solution

after passing through a magnetic field, the magnitude of its mass and velocity of the particle remains same, so its energy does not change that is kinetic energy will remain T.
Hence (C) is correct answer
Question 9
1 / -0

Two parallel wires in free space are $$10\ cm$$ apart and each carries a current of $$10A$$ in the same direction. The force exerted by one wire on other per metre of length of the wire is

A
$$2\times 10^{-6}N$$

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

C
$$2\times 10^{-3}N$$

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

Solution

Given,
$$i_1=i_2=10A$$
$$d=10cm=0.1m$$
$$l=1m$$
The force exerted by one wire on other is given by
$$F=\dfrac{\mu_0 i_1 i_2 l}{2\pi d}$$
$$F=\dfrac{4\pi \times 10^{-7}\times 10\times 10\times 1}{2\pi \times 0.1}=2\times 10^{-4}N$$
The correct option is B.
Question 10
1 / -0

$${\vec B_1},{\vec B_2}$$ and $${\vec B_3}$$ are the magnetic field due to $$I_1 , I_2$$ and $$I_3$$. Ampere's circuital law is given by $$\oint {\vec B} .d\vec l = {\mu _0}I$$, here $$\vec B$$ is_____.

A
$$\vec B = {\vec B_1} - {\vec B_2}$$

B
$$\vec B = {\vec B_1} + {\vec B_2} + {\vec B_3}$$

C
$$\vec B = {\vec B_1} + {\vec B_2} - {\vec B_3}$$

D
$$\vec B = {\vec B_1} - {\vec B_3} $$

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Moving Charges and Magnetism Test - 60

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Question 1

$$2.4 \times 10^{-10} N$$

$$2.403 \times 10^{-12} N$$

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

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

Solution

Question 2

remains constant

increases

decreases

becomes zero

Solution

Question 3

$$2 \times 10^{-7}\ N$$

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

$$0.5\ N$$

$$10^{7}\ N$$

Solution

Question 4

$$0.327 Weber/m^2$$

$$3.27 Weber/m^2$$

$$32.7 Weber/m^2$$

none of these

Solution

Question 5

$$10\ MeV$$

$$20\ MeV$$

$$30\ MeV$$

$$40\ MeV$$

Solution

Question 6

electron

proton

$$He^+$$

$$Li^+$$

Solution

Question 7

Attract to each other

Repel to each other

Both (1) & (2)

They will not experience any force

Solution

Question 8

$$3T$$

$$2T$$

$$T$$

$$4T$$

Solution

Question 9

$$2\times 10^{-6}N$$

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

$$2\times 10^{-3}N$$

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

Solution

Question 10

$$\vec B = {\vec B_1} - {\vec B_2}$$

$$\vec B = {\vec B_1} + {\vec B_2} + {\vec B_3}$$

$$\vec B = {\vec B_1} + {\vec B_2} - {\vec B_3}$$

$$\vec B = {\vec B_1} - {\vec B_3} $$

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