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Tutor profile: Samananda K.

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Samananda K.
K2 physics teacher for 5 years plus
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Questions

Subject: Physics (Newtonian Mechanics)

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Question:

A constand decelerating force of $$ 50\ N $$ is applied to a body of mass $$ 20\ kg $$ moving with a speed of $$ 15\ m/s $$. How long will it take to stop?

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Samananda K.
Answer:

We have, Force acting on the body, $$ F = 50\ N $$ Mass of the body, $$ m = 20\ kg $$ Initial velocity of the body, $$ u = 15\ m/s $$ Final velocity of the body, $$ v = 0 $$ By Newton's second law of motion, $$ F = ma $$ $$ \therefore a = \frac {F} {m} = \frac {- 50} {20} = - 2.5\ ms^{-2} $$ Let $$ t $$ be the time taken by the body to stop its motion. By kinematic equation, $$ v = u + at $$ $$ \therefore t = \frac{ v - u} {a} = \frac {0 - 15} {- 2.5 } = 6\ s $$ Thus, the body takes $$ 6\ s $$ to stop its motion.

Subject: Physics (Electricity and Magnetism)

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Question:

An electron emitted by a heated cathode and accelerated through a potential diffrence of $$ 2.0\ kV $$ enters a region with uniform magnetic field of $$ 0.15\ T $$. Describe the nature of the trajectory of the electron if the field is transverse to its initial velocity.

Inactive
Samananda K.
Answer:

We have, Potential difference, $$ V = 2.0\ kV = 2.0 \times 10^3\ V $$ Magnetic field, $$ B = 0.15\ T $$ Charge of an electron, $$ e = 1.6\times 10^{-19}\ C $$ Mass of an electron, $$ m = 9.1 \times 10^{-31}\ kg $$ Kinetic energy of the electron is given by $$ eV = \frac {1} {2} m v^2 $$ $$ \therefore v = \sqrt{\frac{2eV} {m}} $$ ................. (1) which is expression for velocity of the electron. The magnetic force on the electron provides the necessary centripetal force required to describe a circular path. So, $$ BeV = \frac{mv^2} {r} $$ The radius of the circular path described by the electron is $$ r = \frac {mv} {Be} $$ ................... (2) Using eq.(1) in eq.(2), we get $$ r = \frac{m} {Be} \left[\frac{2eV} {m} \right]^{1/2} $$ $$ = \frac{9.1 \times 10^{-31}} {0.15 \times 1.6 \times 10^{-19}} \times \left [\frac {2 \times 1.6 \times 10^{-19} \times 2 \times 10^3} {9.1 \times 10^{-31}} \right]^{1/2} $$ $$ = 0.5\ mm $$ Thus, the electron describes a circular trajectory of radius $$ 0.5\ mm$$ along the direction of the mgnetic field.

Subject: Physics

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Question:

A plane electromagnetic wave travels in vacuum along $$ z-direction $$. What can you say about the direction of its electric and magnetic field vectors? If the frequency of the wave is $$ 30\ MHz $$, what is its wavelength?

Inactive
Samananda K.
Answer:

The electric field ($$ \mathbf {E}$$) and magnetic field ($$ \mathbf{B} $$) are mutually perpendicular to one another and lies in the $$ xy-plane $$. The frequency of the electromagnetic wave, $$ \nu = 30\ MHz = 30 \times 10^6 s^{-1} $$ Speed of light in vacuum, $$ c = 3 \times 10^{8} m/s $$ The wavelength of the electromagnetic wave is given by $$ \lambda = \frac {c} {\nu} = \frac {3 \times 10^{8}} {30 \times 10^6} m = 10\ m $$

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