Moving Charges and Magnetism Test

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Moving Charges and Magnetism Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

A and B are two like parallel forces. A couple of moment H lies in the plane of A and B and is contained with them. The resultant of A and B after combining is displaced through a distance

A
$$\dfrac{2H}{A-B}$$

B
$$\dfrac{H}{A+B}$$

C
$$\dfrac{H}{2(A+B)}$$

D
$$\dfrac{H}{(A-B)}$$

Solution

Given
$$A$$ and $$B$$ are two like parallel forces.
A couple of moment $$H$$ liesin the Plane of $$A$$ and $$B$$ and is contained with them.
$$\left( A+B \right) d=H$$
So the resultant of $$A$$ and $$B$$after combining is displaced through a distance
Hence the correct answer is $$d=$$ $$\cfrac { H }{ \left( A+B \right) } $$
Question 2
1 / -0

Two long conductors, separated by a distance d carry current $$I_1$$ and $$I_2$$ in the same direction. They exert a force F on each other. Now the current in one of them is increased to two times and its direction is reversed. the distance is also increased to 3d. The new value of the force between them is

A
$$- \displaystyle \frac{2F}{3}$$

B
$$\displaystyle \frac{F}{3}$$

C
-2 F

D
$$- \displaystyle \frac{F}{3}$$

Solution

$$F = \displaystyle \left ( \dfrac{\mu_0 I}{2 \pi r} \right ) l I_2$$ When $$I, = -2 I,$$ and $$r = 3r$$, then

$$\displaystyle F' = - \dfrac{\mu_0 2I_1 lI_2}{2 \pi . 3r}$$

$$= \displaystyle \left ( \dfrac{\mu_0 I_1 I_2 l}{2 \pi r} \right ) \left ( -\dfrac{2}{3} \right ) = - \dfrac{2}{3} F$$
Question 3
1 / -0

The AC voltage across a resistance can be measured using a

A
moving magnet galvanometer

B
moving coil galvanometer

C
hot wire voltmeter

D
potentiometer

Solution

Hint:
For the AC source, The current and the voltage are not constant.

The explanation for the correct answer:
$$\bullet$$ Galvanometer is a device that measures instantaneous value of current in the circuit. Thus, if we measure current in the circuit containing AC source using Galvanometer, the reading of Galvanometer will not remain constant. Thus, we cannot measure AC voltage using a Galvanometer. Options A and B are incorrect.
$$\bullet$$ Hotwired devices are those which can measure any quantity for both AC and DC supply. Hotwired voltmeter can measure the voltage of DC as well as AC source. Option C is correct.
$$\bullet$$ Potentiometers are the device that can measure EMF or internal resistance for any source. It can not measure the potential differences across any resistor. Option D is incorrect.

Thus, only Option C is correct.
Question 4
1 / -0

A particle of mass $$\mathrm{M}$$ and charge $$\mathrm{Q}$$ moving with velocity $$\vec{V}$$ describes a circular path of radius $$\mathrm{R}$$ when subjected to a uniform transverse magnetic field of induction B. The work done by the field when the particle completes one full circle is

A
$$(\displaystyle \frac{\mathrm{M}\mathrm{v}^{2}}{\mathrm{R}})2\pi \mathrm{R}$$

B
zero

C
$$\mathrm{B}2\pi \mathrm{R}$$

D
$$\mathrm{B}\mathrm{v}2\pi \mathrm{R}$$

Solution

The work done by the magnetic field is zero and also here the work done in a one complete circular path is zero because the particle displacement is zero.
Question 5
1 / -0

Wires 1 and 2 carrying currents $$i_1$$ and $$i_2$$ respectively are inclined at an angle $$\theta$$ to each other. What is the force on a small element d$$l$$ of wire 2 at a distance of r from wire 1 (as shown in figure) due to the magnetic field of wire 1?

A
$$\displaystyle \frac{\mu_0 }{2 \pi r} i_1 i_2 d l tan \theta$$

B
$$\displaystyle \frac{\mu_0 }{2 \pi r} i_1 i_2 d l sin \theta$$

C
$$\displaystyle \frac{\mu_0 }{2 \pi r} i_1 i_2 d l cos \theta$$

D
$$\displaystyle \frac{\mu_0 }{4 \pi r} i_1 i_2 d l sin \theta$$

Solution
Question 6
1 / -0

A particle of mass M and charge Q moving with velocity $$\vec v$$ describe a circular path of radius R when subjected to a uniform transverse magnetic field of induction B. The work done by the field when the particle completes one full circle is

A
$$\displaystyle \left ( \frac{Mv^2}{R} \right ) 2 \pi R$$

B
$$zero$$

C
$$BQ2 \pi R$$

D
$$BQv2 \pi R$$

Solution

Upon completing a full circle net displacement is 0.
Work done by the magnetic field is 0 because the net displacement caused by the magnetic field is 0.
Question 7
1 / -0

A charged particle moves through a magnetic field perpendicular to its direction. Then

A
kinetic energy changes but the momentum is constant

B
The linear momentum changes but the kinetic energy is constant

C
Both linear momentum and kinetic energy of the particle are not constant

D
Both linear momentum and kinetic energy of the particle are constant

Solution

When a charge particle enters a magnetic field at a direction perpendicular to the direction of motion, the path of the motion is circular. In circular motion the direction of velocity changes at every point (the magnitude remains constant). Therefore, the tangential momentum will change at every point. But kinetic energy will remain constant as it is given by $$\frac{1}{2} mv^2$$ and $$v^2$$ is the square of the magnitude of velocity which does not change.
In this question, we could not infer that where the momentum mentioned here is tangential or angular.
If it is angular momentum, then (d) is the correct option.
Question 8
1 / -0

A particle of charge $$-16\times 10^{-18}$$ coulomb moving with velocity 10 $$\mathrm{m}\mathrm{s}^{-1}$$ along the x-axis enters a region where a magnetic field of induction $$\mathrm{B}$$ is along the $$\mathrm{y}$$ -axis, and an electric field of induction $$\mathrm{B}$$ is along the $$\mathrm{y}$$-axis, and an electric field of magnitude $$10^{4}\mathrm{V}/\mathrm{m}$$ is along the negative $$\mathrm{z}$$-axis. lf the charged particle continues moving along the $$\mathrm{x}$$-axis, the magnitude of $$\mathrm{B}$$ is

A
$$10^{3}\mathrm{W}\mathrm{b}/\mathrm{m}^{2}$$

B
$$10^{5}\mathrm{W}\mathrm{b}/\mathrm{m}^{2}$$

C
$$10^{16}\mathrm{W}\mathrm{b}/\mathrm{m}^{2}$$

D
$$10^{-3}\mathrm{W}\mathrm{b}/\mathrm{m}^{2}$$

Solution

Particle travels along $$x$$ axis
Hence $${ V }_{ y }={ V }_{ z }=0$$
Electric feild is along the negative $$z$$ axis
$${ E }_{ x }={ E }_{ y }=0$$
Therefore net force on the particle $$\vec { F } =q\left( \vec { E } +V\times \vec { B } \right) $$
Resolve the motion along three coordinate axis
$${ a }_{ x }=\cfrac { { F }_{ x } }{ m } =\cfrac { q }{ m } \left( { E }_{ x }+{ V }_{ y }{ B }_{ z }-{ V }_{ x }{ B }_{ y } \right) \\ { a }_{ y }=\cfrac { { F }_{ x } }{ m } =\cfrac { q }{ m } \left( { E }_{ y }+{ V }_{ z }{ B }_{ x }-{ V }_{ xz } \right) \\ { a }_{ z }=\cfrac { { F }_{ x } }{ m } =\cfrac { q }{ m } \left( { E }_{ z }+{ V }_{ x }{ B }_{ y }-{ V }_{ y }{ B }_{ x } \right) $$
Since,
$${ V }_{ y }={ V }_{ z }=0\\ { E }_{ x }={ E }_{ y }=0\\ { B }_{ x }={ B }_{ z }=0$$
$${ a }_{ x }={ a }_{ y }=0\\ { a }_{ z }=\cfrac { q }{ m } \left( -{ E }_{ z }+{ V }_{ x }{ B }_{ y } \right) $$
Again $${ a }_{ z }=0$$ , as the particle transverse through the region undeflected
$${ E }_{ z }={ V }_{ x }{ B }_{ y }$$
$${ B }_{ y }=\cfrac { { E }_{ z } }{ { V }_{ x } } =\cfrac { { 10 }^{ 4 } }{ 10 } ={ 10 }^{ 3 }wb/m^{ 2 }$$
Question 9
1 / -0

A charged particle moves through a magnetic field perpendicular to its direction. Then

A
the momentum changes but the kinetic energy is constant

B
both momentum and kinetic energy of the particle are not constant

C
both, momentum and kinetic energy of the particle are constant

D
kinetic energy changes but the momentum is constant

Solution

The work done is the change in kinetic energy (KE). Since the work done by the magnetic field is zero. So, KE $$=$$ constant.
But the particle is moving through a magnetic so velocity changes it means momentum also changes.
Question 10
1 / -0

Two long conductors, separated by a distance $$d$$ carry currents $$I_1$$ and $$I_2$$ in the same direction. They exert a force $$F$$ on each other. Now the current in one of them is increased to two times and its direction is reversed. The distance is also increased to $$3d$$. The new value of the force between them is:

A
$$-2F$$

B
$$\dfrac{F}{3}$$

C
$$\dfrac{-2F}{3}$$

D
$$\dfrac{-F}{3}$$

Solution

$$\text{Magnetic field due to a wire at a distance}\ d\ \text{is}\ B=\dfrac{\mu_0 I_1}{2\pi d}$$
$$\text{Force on other wire due to this magnetic field of length} \ l \ \text{is}\ F=BI_2l=\dfrac{\mu_0 I_1}{2\pi d}I_2 l$$
$$\text{Therefore force per unit length}=F/l=\dfrac{\mu_0 I_1I_2}{2\pi d}$$
$$\Rightarrow F/l\propto \dfrac{I_1I_2}{d}$$
$$\text{When current becomes two times and distance is increased to three times,}$$
$$F_1=\dfrac{2F}{3}$$
$$\text{Since the direction of current is also reversed in one of them, the direction of force also reverses.}$$
$$\Rightarrow F^{'}=\dfrac{-2F}{3}$$

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Re-Attempt Weekly Quiz Competition

Moving Charges and Magnetism Test - 20

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Moving Charges and Magnetism Test - 20

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Question 1

$$\dfrac{2H}{A-B}$$

$$\dfrac{H}{A+B}$$

$$\dfrac{H}{2(A+B)}$$

$$\dfrac{H}{(A-B)}$$

Solution

Question 2

$$- \displaystyle \frac{2F}{3}$$

$$\displaystyle \frac{F}{3}$$

-2 F

$$- \displaystyle \frac{F}{3}$$

Solution

Question 3

moving magnet galvanometer

moving coil galvanometer

hot wire voltmeter

potentiometer

Solution

Question 4

$$(\displaystyle \frac{\mathrm{M}\mathrm{v}^{2}}{\mathrm{R}})2\pi \mathrm{R}$$

zero

$$\mathrm{B}2\pi \mathrm{R}$$

$$\mathrm{B}\mathrm{v}2\pi \mathrm{R}$$

Solution

Question 5

$$\displaystyle \frac{\mu_0 }{2 \pi r} i_1 i_2 d l tan \theta$$

$$\displaystyle \frac{\mu_0 }{2 \pi r} i_1 i_2 d l sin \theta$$

$$\displaystyle \frac{\mu_0 }{2 \pi r} i_1 i_2 d l cos \theta$$

$$\displaystyle \frac{\mu_0 }{4 \pi r} i_1 i_2 d l sin \theta$$

Solution

Question 6

$$\displaystyle \left ( \frac{Mv^2}{R} \right ) 2 \pi R$$

$$zero$$

$$BQ2 \pi R$$

$$BQv2 \pi R$$

Solution

Question 7

kinetic energy changes but the momentum is constant

The linear momentum changes but the kinetic energy is constant

Both linear momentum and kinetic energy of the particle are not constant

Both linear momentum and kinetic energy of the particle are constant

Solution

Question 8

$$10^{3}\mathrm{W}\mathrm{b}/\mathrm{m}^{2}$$

$$10^{5}\mathrm{W}\mathrm{b}/\mathrm{m}^{2}$$

$$10^{16}\mathrm{W}\mathrm{b}/\mathrm{m}^{2}$$

$$10^{-3}\mathrm{W}\mathrm{b}/\mathrm{m}^{2}$$

Solution

Question 9

the momentum changes but the kinetic energy is constant

both momentum and kinetic energy of the particle are not constant

both, momentum and kinetic energy of the particle are constant

kinetic energy changes but the momentum is constant

Solution

Question 10

$$-2F$$

$$\dfrac{F}{3}$$

$$\dfrac{-2F}{3}$$

$$\dfrac{-F}{3}$$

Solution

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