Moving Charges and Magnetism Test

Result

Moving Charges and Magnetism Test - 75

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A photon of energy E ejects a photoelectron from a metal surface whose work function is $$W_0.$$ If this electron enters into a uniform magnetic field of induction B in a direction perpendicular to the field and describes a circular path of radius r, then the radius r is given by (in the usual notation)

A
$$\displaystyle\sqrt \frac{{2\, m \, (E \, - \, W_0)}}{eB}$$

B
$$\sqrt{2 \,m \, (E \, -\, W_0) \, {eB}}$$

C
$$\displaystyle \frac{\sqrt{2\, m \, (E \, - \, W_0)}}{mB}$$

D
$$\displaystyle \frac{\sqrt{2\, m \, (E \, - \, W_0)}}{eB}$$

Solution

From Einstein's equation,
energy of incident photon
$$E \, = \, W_0 \, + \, \displaystyle \frac{1}{2} \, mv^2$$
We get, velocity of photoelectron
$$\displaystyle v = \sqrt{\frac {2 \, (E \, - \, W_0)}{m}} $$
A charged particle placed in uniform magnetic field experience a magnetic force which makes the charge to trace a circular path.
Centripetal force $$F \, = \, \displaystyle \frac{mv^2}{r} $$

Magnetic force $$F_m = evB$$

or $$evB \, = \, \displaystyle \frac{mv^2}{r}$$

or $$r \, = \, \displaystyle \frac{mv}{eB}$$

or $$r \, = \, \displaystyle \frac{m \, \sqrt{\displaystyle \frac{2 \, (E \, -\, W_0)}{m}}}{eB}$$

$$\implies \ r \, = \, \displaystyle \frac{\sqrt{2\, m (\, E \, - \, W_0)}}{eB}$$
Question 2
1 / -0

A magnetic dipole $$\overrightarrow {\rm{M}} {\rm{ = }}\left( {{\rm{A}}\widehat {\rm{i}}{\rm{ + B}}\widehat {\rm{j}}} \right){\rm{J/Wb}}$$ is placed in magnetic field. $$\overrightarrow {\rm{B}} {\rm{ = }}\left( {{\rm{C}}{{\rm{x}}^{\rm{2}}}\widehat {\rm{i}}{\rm{ + D}}{{\rm{y}}^{\rm{2}}}\widehat {\rm{j}}} \right)$$ Wb in XY plane at $$\overrightarrow {\rm{r}} {\rm{ = }}\left( {{\rm{E}}\widehat {\rm{i}}{\rm{ + F}}\widehat {\rm{j}}} \right){\rm{m}}$$. Then force experienced by the dipole is:

A
$${\rm{2ACE}}\widehat {\rm{i}}{\rm{ + 2BDF}}\widehat {\rm{j}}\left( {\rm{N}} \right)$$

B
$${\rm{2ACE}}\widehat {\rm{i}}\left( {\rm{N}} \right)$$

C
0

D
$${\rm{ACE}}\widehat {\rm{i}}{\rm{ + BDF}}\widehat {\rm{j}}\left( {\rm{N}} \right)$$

Solution
Question 3
1 / -0

In an experiment, setup $$A$$ consists of two parallel wires which carry currents in opposite directions as shown in the figure. $$A$$ second setup $$B$$ is identical to setup $$A$$, except that there is a metal plate between the wires.
Let $$F_A$$ and $$F_B$$ be the magnitude of the force between the two wires in setup $$A$$ and $$B$$, respectively, then which of the following condition is correct:

A
$$F_A> F_B \neq 0$$

B
$$F_A< F_B$$

C
$$F_A = F_B \neq 0$$

D
$$F_A > F_B = 0$$

Solution

Placing a metal plate between the wires in setup $$B$$ does not affect the magnetic force exerted between the wires since,
$$F=\dfrac{\mu_0}{2\pi}-\dfrac{I_1I_2}{r^2}$$
is independent of dielectric constant (which changes due to metal sheet.)
Question 4
1 / -0

An electron emitted from a hot filament is accelerated through a potential difference of 2.88 kV and enters a region of a uniform magnetic field of $$0.\pi$$ at an angle of $$30^{o}$$ with the field. Take the mass of an electron $$9 \times 10^{-31}$$ kg. The trajectory of the electron is

A
Circle of radius 1.6 mm

B
Circle of radius 0.9 mm

C
Helix of radius 0.9 mm

D
Helix of radius 1.6 mm

Solution
Question 5
1 / -0

An electron accelerated through a potential difference $$V$$ passes through a uniform transverse magnetic field and experiences a force $$F$$. If the accelerating potential is increased to $$2V$$, the electron in the same magnetic field will experience a force :

A
$$F$$

B
$$F/2$$

C
$$\sqrt {2} F$$

D
$$2F$$

Solution

force is given as:

$$\quad F=Bqv\quad $$

The kinetic energy of an electron after passing through a potential difference of V volt is given as

But $$\cfrac { 1 }{ 2 } m{ v }^{ 2 }=eV$$ or

$$V=\sqrt { \cfrac { 2eV }{ m } } $$

$$F=Bq\sqrt { \cfrac { 2eV }{ m } } \Rightarrow F\propto \sqrt { V } ;F'\propto \sqrt { 2V } ;$$

$$\cfrac { F' }{ F } =\sqrt { 2 } \quad or\quad F'=\sqrt { 2 } F$$
Question 6
1 / -0

The magnetic field due to current flowing in a ling straight conductor is directly proportional to the current and inversely proportional to the distance of the point of observation from the conductor. What is this law known as?

A
Blonde-Rey law

B
Biot-Savart's law

C
Beer-Lambert law

D
Ampere's law

Solution
Question 7
1 / -0

An electron gun G emits electrons of energy $$2$$ keV travelling in the positive x-direction. The electrons are required to hit the spot S where GS$$=0.1$$m, and the line GS makes an angle $$60^o$$ with the x-axis, as shown. A uniform magnetic field B parallel to GS exists in the region outside the electron gun. Find the minimum value of B needed to make the electrons hit S.

A
$$1.7\times 10^{-3}$$T

B
$$5.7\times 10^{-3}$$T

C
$$3.7\times 10^{-3}$$T

D
$$4.7\times 10^{-3}$$T

Solution
Question 8
1 / -0

Two thin, long parallel wires separated by a distance by a distance $$d$$ carry a current of $$iA$$ in the same direction. They will

A
Attract each other with a force per unit length of $$\mu_{0}i^{2}/(2\pi d)$$

B
Repel each other with a force per unit length of $$\mu_{0}i^{2}\ /(2\pi d)$$

C
Attract each other with a force per unit length of $$\mu_{0}i^{2}(2\pi d^{2})$$

D
Repel each other with a force per unit length of $$\mu_{0}i^{2}/(2\pi d^{2})$$

Solution
Question 9
1 / -0

An electron is projected with uniform velocity along the axis of a current carrying long solenoid. Which of the following is true?

A
The electron will be accelerated along the axis.

B
The electron path will be circular about the axis.

C
The electron will experience a force at 45 to the axis and hence execute a helical path.

D
The electron will continue to move with uniform velocity along the axis of the solenoid.

Solution

Hint:- If a charge $$q$$ is moving with velocity $$v$$ in a Magnetic Field $$B$$, such that angle between velocity of charge and Magnetic Field is $$\theta$$. Then Force acting on charge is given by,
$$F = qvBsin\theta$$

Step 1: Note the given values and calculate force on an electron in magnetic field.
$$\bullet$$ It is given that an electron is projected with uniform velocity along the axis of current carrying long solenoid. Let charge on electron is $$e$$ and uniform velocity of electron is $$v$$. Also let at any point on axis of solenoid Magnetic Field is $$B$$.
$$\bullet$$ We know that Magnetic Field on axis of solenoid is always parallel to the axis. And since electron is moving along axis of solenoid, we have $$\theta = 0°$$ or $$\theta = 180°$$. In both the cases value of $$\sin\theta$$ is zero.
$$\bullet$$ Thus net force acting on electron, that is
$$F = qvBsin\theta = 0$$

Step 2: Explanation for options:-
$$\bullet$$ Since net force on electron is zero, by Newton's Second Law its acceleration will be zero. Option A is incorrect.
$$\bullet$$ Also since net force on electron is zero, its direction of motion will not change. So Option B is incorrect.
$$\bullet$$As theta is zero not 45 so option C is incorrect.
$$\bullet$$ As net force is zero so net acceleration is zero. So the electron will continue to move along the axis with same velocity.
Thus, only Option D is correct.
Question 10
1 / -0

Two charged particles M and N are projected with the same velocity in a uniform magnetic field as shown. Then, M and N, respectively, can be

A
An $$\alpha $$-particle and a proton

B
A deutron and a proton

C
A deutron and an $$\alpha$$- particle

D
A proton and an $$\alpha$$- particle