Question

need before 12:00 noon central time july 30, 2020

Questions: R 1) Two resistors are connected to each other in parallel, one resistor has a known value of R, the resistance of the second one is unknown. If the equivalent of two resistors in parallel is equal to what is the resistance of the second one? A)R B) 2R C) D) E) 4R EINE 2) Which one of the following sentences is correct about the velocity selector? A) The velocity of charged particles leaving a velocity selector are the same for all particles independent value of charge and value of mass. B) Velocity of a charged particle leaving a velocity selector is equal to v = B/E C) The velocity of a charge leaving the selector is independent of electric field. D) A velocity selector is made up of magnetic field and electric field which are parallel with respect to each other. E) The velocity of charge leaving the selector is independent of the magnetic field. 3) A charged particle is moving with velocity of Vin a magnetic field of B, which one of the followings is correct: A) The direction of force F on the charge is parallel to magnetic field B B) The direction of force F on the charge is parallel to velocity direction V C) The force is maximized when velocity direction and magnetic field are parallel D) The force F is perpendicular (normal) to both velocity Vand magnetic field B E) The direction of foron positive charge or negative charge would the same

Answer 1

Solution of 1:

We are given two resistors are connected in parallel combination, one resistor has a known value of , the value of second resistor $R_{x}$ is unknown, and the equivalent resistance of parallel combination is
R/2
:

2 = R R R.

$\therefore \frac{1}{R_{x}}=\frac{2}{R}-\frac{1}{R}$

${\color{Blue} \therefore \frac{1}{R_{x}}=\frac{1}{R}}$

${\color{Blue} \therefore R_{x}={R}}$

Therefore, the correct option is (A) ${\color{Blue} {R}}$

Solution of 2:

The magnetic force $F_{m}$ is given as:

$F_{m}=q(\overrightarrow{v}\times \overrightarrow{B})$

The electric force $F_{e}$ is given as:

$F_{e}=qE$

The total force on the charged particle is given as:

$F=F_{e}+F_{m}$

$\therefore F=qE+q(\overrightarrow{v}\times \overrightarrow{B})$

$\therefore F=qE+q(vB\ sin\theta)$

Here, is the charge of the particle, is the electric field, is the velocity of the particle, is the magnetic field, and $\theta$ is the angle between velocity vector and magnetic field vector.

Therefore, the magnetic force depends on the velocity vector $\overrightarrow{v}$ . When $sin90\degree=1$ , the magnetic field vector and velocity vector is perpendicular to each other and magnetic force is maximum. When $sin0\degree=0$ , the magnetic field vector and velocity vector is parallel to each other and magnetic force is zero.

When electric force $F_{e}$ and magnetic force $F_{m}$ are equal, and $sin90\degree=1$ , the velocity is given as:

$\therefore F_{e}=F_{m}$

$\therefore qE=qvB$

${\color{Blue} \therefore v=\frac{E}{B}}$

This is used to select a charged particle of a particular velocity from charges moving with different speeds which is called the velocity selector and therefore, the correct option is (B) Velocity of a charged particle leaving a velocity selector is equal to ${\color{Blue} v=E/B}$ .

Solution of 3:

Since velocity vector $\overrightarrow{v}$ , magnetic field
B
, and electric field
E
are always perpendicular to each other, therefore, the correct option is (E) The direction of force on positive or negative charge would be the same.