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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.

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.

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Samananda K.

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.

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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