Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 26

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Question 1
1 / -0

A magnetic field cannot exert any force on a

A
Moving magnet

B
Stationary magnet

C
Moving charge

D
Stationary charge

Solution

Magnetic fields are produced by magnetic bodies ( Moving or stationary ) . Moving charges produce electromagnetic fields. Stationary charges produce electric fields alone.
Question 2
1 / -0

A magnetic field can exert force on a

A
stationary magnet

B
moving charge

C
moving magnet

D
all of these

Solution

Hint:

The magnetic field is a vector quantity that describes the magnetic influence on moving charges, electric currents, and magnetic materials.

Explanation:

$$\bullet$$ Magnetic field exerts a force on a magnet irrespective of its motion, which is given by

$$F=mB$$

Where $$m$$ is the magnetic dipole moment of a magnet, $$B$$ is Magnetic Field.

$$\bullet$$ Magnetic field exerts a force on moving charges given by,

$$\overrightarrow{F}=q\left( \overrightarrow{v}\times \overrightarrow{B} \right)$$

Where $$q$$ is charge, $$\overrightarrow{v}$$ is velocity of charge, $$\overrightarrow{B}$$ is magnetic Field

Thus, Magnetic Field exerts a force on stationary magnets, moving charges, moving magnets.
Question 3
1 / -0

The correct expression for Ampere's law is :

A
$$\int B.dl=\sum i$$

B
$$\int B.dl=\dfrac {1}{\sum i}$$

C
$$\int B.dl=\mu_0\sum i$$

D
$$\int B.dl=\dfrac {\sum i}{\mu_0}$$

Solution

Amper's law is $$\oint \vec{B}.\vec{dl} = \mu_0 \sum i$$ where $$i$$ is the sum of enclosed currents by the loop.
Question 4
1 / -0

If a charged particle moves through a magnetic field, then effect of the field is to change the particle's

A
speed

B
energy

C
acceleration

D
direction of motion

Solution

Magnetic force on a moving charge placed in a magnetic field is given by,
$$F = q(\mathbf{V} X \mathbf{B})$$
we can see from the equation that magnetic force is cross product of magnetic field strength and velocity ,hence the direction of force will be perpendicular to both of them.
And it will try to change the direction of motion.
Question 5
1 / -0

If a 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
both inside and outside the pipe

D
neither inside not outside the pipe

Solution

For a loop inside the hollow portion of the conductor, the current enclosed is zero until the loop is on the surface or inside the material of the conductor.
∴Binside=0
For any loop outside, current enclosed by the amperian loop is not zero, hence magnetic field is non-zero.
Question 6
1 / -0

Two parallel wires carrying current 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

Force on a current carrying conductor is $$F=\int i (\vec{dl} \times \vec{B})$$
Magnetic field due to one wire at the location of the second wire ( other one ) is $$B_1= \frac{\mu_0 i_1}{2\pi d}$$
Forceon the 2nd wire due to $$B_1$$ is $$F_{21}= i_2l_2B_1= i_2l_2 \dfrac{\mu_0 i_1}{2\pi d} =\dfrac{\mu_0 i_1i_2l_2}{2\pi d}$$
Force per unit length for wire $$l_2$$ is $$F_{ per\: unit\: length}=\dfrac{F_{21}}{l_2}$$
The force is attractive.
Question 7
1 / -0

Suppose that a proton traveling in vaccum with velocity $$u_1$$ at right angles to a uniform magnetic field experiences twice the force that an $$\alpha$$ particle experiences when it is traveling along the same path with velocity $$u_2$$.The ratio $$\left (\dfrac {u_1}{u_2}\right )$$ is

A
$$\dfrac {1}{2}$$

B
$$1$$

C
$$2$$

D
$$4$$

Solution

$$F_p= q_pu_1B$$
$$F_{\alpha} =q_{\alpha}u_2B$$
Given, $$F_p = 2F_{\alpha}$$
$$ \therefore q_pu_1B = 2q_{\alpha}u_2B$$
$$ \therefore \dfrac{u_1}{u_2} = 2\dfrac{q_{\alpha}}{q_p}=2 \times 2 =4$$ since $$q_{\alpha}=2q_p$$
Question 8
1 / -0

The correct expression for Lorentz force is

A
$$q[\vec E+(\vec B\times \vec V)]$$

B
$$q[\vec E+(\vec V\times \vec B)]$$

C
$$q(\vec V\times \vec B)$$

D
$$q\vec E$$

Solution

In electromagnetic field both electric and magnetic forces are experienced by the charge,
F = electric force + magnetic force
$$F = qE + q(\vec v \times B)$$
$$F = q[E + (\vec v \times B)]$$
Question 9
1 / -0

Which of the following expressions are applicable to the moving coil galvanometer?

A
$$\displaystyle \overrightarrow{F_m}=q(\overrightarrow {V}\times\overrightarrow{B})$$

B
$$\displaystyle B=B_0\tan\theta$$

C
$$\displaystyle \overrightarrow {\tau}=\overrightarrow{M}\times\overrightarrow{B}$$

D
none of these

Solution

Torque $$\overrightarrow{\tau} = \overrightarrow{M} \times \overrightarrow{B}$$
Torque is applicable to the moving coil galvanometer.
Question 10
1 / -0

A moving coil galvanometer is based on the

A
heating effect of current

B
magnetic effect of current

C
chemical effect of current

D
peltier effect of current

Solution

Torque acts on the coil due to magnetic field when current flows in it.
Hence, it is based on magnetic effect of current.

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Moving Charges and Magnetism Test - 26

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Moving Charges and Magnetism Test - 26

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Question 1

Moving magnet

Stationary magnet

Moving charge

Stationary charge

Solution

Question 2

stationary magnet

moving charge

moving magnet

all of these

Solution

Question 3

$$\int B.dl=\sum i$$

$$\int B.dl=\dfrac {1}{\sum i}$$

$$\int B.dl=\mu_0\sum i$$

$$\int B.dl=\dfrac {\sum i}{\mu_0}$$

Solution

Question 4

speed

energy

acceleration

direction of motion

Solution

Question 5

only inside the pipe

only outside the pipe

both inside and outside the pipe

neither inside not outside the pipe

Solution

Question 6

potential difference between them

mutual inductance between them

electric forces between them

magnetic forces between them

Solution

Question 7

$$\dfrac {1}{2}$$

$$1$$

$$2$$

$$4$$

Solution

Question 8

$$q[\vec E+(\vec B\times \vec V)]$$

$$q[\vec E+(\vec V\times \vec B)]$$

$$q(\vec V\times \vec B)$$

$$q\vec E$$

Solution

Question 9

$$\displaystyle \overrightarrow{F_m}=q(\overrightarrow {V}\times\overrightarrow{B})$$

$$\displaystyle B=B_0\tan\theta$$

$$\displaystyle \overrightarrow {\tau}=\overrightarrow{M}\times\overrightarrow{B}$$

none of these

Solution

Question 10

heating effect of current

magnetic effect of current

chemical effect of current

peltier effect of current

Solution

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