Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 69

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Weekly Quiz Competition

Question 1
1 / -0

An electron is moving on a circular path of radius $$r$$ with speed $$v$$ in a transverse magnetic field $$B$$. $$e/m$$ for it will be

A
$$\dfrac {v}{rB}$$

B
$$\dfrac {B}{rv}$$

C
$$Bvr$$

D
$$\dfrac {vr}{B}$$

Solution

The radius is given by $$r=\dfrac {mv}{eB}\Rightarrow \dfrac {e}{m}=\dfrac {v}{rB}$$
Question 2
1 / -0

When a magnetic field is applied in a direction perpendicular to the direction of cathode rays, then their

A
Energy increase

B
Energy decrease

C
Momentum increase

D
Momentum and energy remain unchanged

Solution

Since force is perpendicular to direction of motion, energy and magnitude of momentum remains constant.
Question 3
1 / -0

A particle is moving in a uniform magnetic field, then

A
It momentum changes but total energy remains the same

B
Both momentum and the total energy the same

C
Both will change

D
Total changes but momentum remains the same

Solution

The charge particle moving in a magnetic field does not gain energy. However the direction of its velocity charges continuously. Hence momentum charges.
Question 4
1 / -0

A small cylindrical soft iron piece is kept in a galvanometer so that

A
A radial uniform magnetic fields is produced

B
A uniform magnetic fields is produced

C
There is a steady deflection of the coil

D
All of these

Solution

A soft iron core is used in a moving coil galvanometer. The soft iron core attracts the magnetic lines of force and hence the strength of the magnetic field increases if we use soft iron core. Thus the sensitivity of galvanometer increases. Also the use of soft iron core makes the magnetic field radial (i.e the plane of the coil will be always parallel to the direction of magnetic field).
Question 5
1 / -0

Motion of a moving electrons is not affected by

A
An electric field applied in the direction of motion

B
Magnetic field applied in the direction of motion

C
Electric field applied perpendicular to the direction of motion

D
Magnetic field applied perpendicular to the direction of motion

Solution

When field is parallel to the direction of motion of charge magnitude force on it is zero.
$$F = -e(\vec v \times \vec B)$$
As $$q$$ and $$B$$ has same direction, so $$\vec v \times \vec B = 0$$
So, $$\vec F =0$$
Question 6
1 / -0

Two long straight parallel conductors separated by a distance of $$0.5\ m$$ carry currents of $$5A$$ and $$8A$$ in the same direction. The force per unit length experienced by each other is

A
$$1.6 \times 10^{-5}\ N$$ (attractive)

B
$$1.6 \times 10^{-5}\ N$$ (repulsive)

C
$$16 \times 10^{-5}\ N$$ (attractive)

D
$$16 \times 10^{-5}\ N$$ (repulsive)

Solution

$$F=10^{-7} \dfrac{2i_1 i_2}{a} =10^{-7} \times \dfrac{2 \times 5 \times 8}{0.5} =1.6 \times 10^{-5}\ N$$
As current in the two wires are in same direction. So, fore will be attractive
Question 7
1 / -0

An electron is travelling in east direction and a magnetic field is applied in upward direction then electron will deflect in

A
South

B
North

C
West

D
East

Solution

By Fleming left hand rule.
The electron moving along East will experience the force towards North direction.

In the presence of an electric field, the charged particles would experience an electric force regardless of their state i.e. stable or in motion.

Another force that the charged particles experience in the absence of an electric field is the magnetic force but they have to be in motion. The magnetic field produces induced electric field.

Since electrons movement is always opposite to current so we can use left hand rule and get the direction of force.
Question 8
1 / -0

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

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

B
$$2 \times 10^{-2}\ N$$, repulsive

C
$$2 \times 10^{-7}\ N$$, attractive

D
$$2 \times 10^{-7}\ N$$, repulsive

Solution

$$F=\dfrac{\mu_0}{4 \pi} \dfrac{2 i_1 i_2}{a}=10^{-7} \times \dfrac{2 \times 10 \times 10}{0.1}=2 \times 10^{-4}\ N$$
Direction of current is same, so force is attractive.
Question 9
1 / -0

A electron $$(q=1.6\times 10^{-19}C)$$ in moving at right angle to the uniform magnetic field $$3.534\times 10^{-5}T$$. The time taken by the electrons to complete a circular orbit is

A
$$2\ \mu s$$

B
$$4\ \mu s$$

C
$$3\ \mu s$$

D
$$1\ \mu s$$

Solution

$$T=\dfrac {2\pi m}{qB}=\dfrac {2\times 3.14\times 9.1\times 10^{-31}}{1.6\times 10^{-19}\times 3.534\times 10^{-5}}$$
$$=1\times 10^{-6} sec =1\ \mu sec$$
Question 10
1 / -0

The magnetic field near a current carrying conductor is given by

A
Coulomb's law

B
Lenz' law

C
Biot-Savart's law

D
Kirchhoff's law

Solution

The formula is $$B = \dfrac{\mu _o}{4\pi} I \dfrac{d\vec l \times \vec r}{r^3}$$
It is Biot-sarvart's law

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Moving Charges and Magnetism Test - 69

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Moving Charges and Magnetism Test - 69

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Question 1

$$\dfrac {v}{rB}$$

$$\dfrac {B}{rv}$$

$$Bvr$$

$$\dfrac {vr}{B}$$

Solution

Question 2

Energy increase

Energy decrease

Momentum increase

Momentum and energy remain unchanged

Solution

Question 3

It momentum changes but total energy remains the same

Both momentum and the total energy the same

Both will change

Total changes but momentum remains the same

Solution

Question 4

A radial uniform magnetic fields is produced

A uniform magnetic fields is produced

There is a steady deflection of the coil

All of these

Solution

Question 5

An electric field applied in the direction of motion

Magnetic field applied in the direction of motion

Electric field applied perpendicular to the direction of motion

Magnetic field applied perpendicular to the direction of motion

Solution

Question 6

$$1.6 \times 10^{-5}\ N$$ (attractive)

$$1.6 \times 10^{-5}\ N$$ (repulsive)

$$16 \times 10^{-5}\ N$$ (attractive)

$$16 \times 10^{-5}\ N$$ (repulsive)

Solution

Question 7

South

North

West

East

Solution

Question 8

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

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

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

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

Solution

Question 9

$$2\ \mu s$$

$$4\ \mu s$$

$$3\ \mu s$$

$$1\ \mu s$$

Solution

Question 10

Coulomb's law

Lenz' law

Biot-Savart's law

Kirchhoff's law

Solution

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