Tag: magnetic fields and electromagnetism

Questions Related to magnetic fields and electromagnetism

Consider two conducting plates $A$ and $B$, between which the potential difference is $5 V$, plate $A$ being at a higher potential. A proton and an electron are released at plates $A$ and $B$ respectively. The two particles then move towards the opposite plates - the proton to plate $B$ and the electron to plate $A$. Which one will have a larger velocity when they reach their respective destination plates?

physics magnetic fields and electromagnetism contact and non-contact forces comparing force in magnetic, electric and gravitational fields identifying forces

  1. Both will have the same velocity

  2. The electron will have the larger velocity

  3. The proton will have the larger velocity

  4. None will be able to reach the destination point

Show answer
Correct Option: B
Explanation:

Between two conducting plates, there is the uniform electric field and therefore the same charge is applicable on both proton and neutron. Since Proton mass is more than that of the electron, it will have less acceleration and hence less speed achieved by it. Therefore electron will reach plate with larger velocity.

$F _g$ and $F _e$ represent gravitational and electrostatic force respectively between electrons situated at a distance $0.1\,m. F _g/F _e$ is of the order

physics magnetic fields and electromagnetism contact and non-contact forces comparing force in magnetic, electric and gravitational fields identifying forces

  1. $10^{-41}$

  2. $10^{-45}$

  3. $10^{40}$

  4. $10^{-42}$

Show answer
Correct Option: D
Explanation:
${ F } _{ y }$ and ${ F } _{ e }$ represent gravitational and electrostatic force respectively between electrons situated at a distance $=0.1m$
let, us take the ratio
$\dfrac { { F } _{ e } }{ { F } _{ g } } =\dfrac { \dfrac { 1 }{ 4\pi { \epsilon  } _{ 0 } } \dfrac { { q }^{ 2 } }{ { d }^{ 2 } }  }{ G\dfrac { { m }^{ 2 } }{ { d }^{ 2 } }  } $
$\dfrac { { F } _{ e } }{ { F } _{ g } } =\dfrac { 1 }{ 4\pi { \epsilon  } _{ 0 } } \times \dfrac { 1 }{ G } \times \dfrac { { q }^{ 2 } }{ { m }^{ 2 } } $
$\dfrac { { F } _{ e } }{ { F } _{ g } } =9\times { 10 }^{ 9 }\times \dfrac { 1 }{ 6.67\times { 10 }^{ -11 } } \times \dfrac { { \left( 1.6\times { 10 }^{ -19 } \right)  }^{ 2 } }{ { \left( 9.1\times { 10 }^{ -31 } \right)  }^{ 2 } } $
$\Rightarrow \dfrac { { F } _{ e } }{ { F } _{ g } } =4.17\times { 10 }^{ 42 }$
or,  $\dfrac { { F } _{ e } }{ { F } _{ g } } $ is of the order of ${ 10 }^{ -42 }$.

Four point charge of $+10^{-7}C, -10^{-7}C, -2\times 10^{-7}C$ and $+2\times 10^{-7}C$ are placed respectively at the corners $A, B, C, D$ of a $0.05\ m$ square. Find the magnitude of the resultant force on the charge at $D$.

physics magnetic fields and electromagnetism contact and non-contact forces comparing force in magnetic, electric and gravitational fields identifying forces

  1. $0.2\ dyne$

  2. $0.2\ newton$

  3. $2\ dyne$

  4. $0.02\ newton$

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Correct Option: A

An $\alpha$-particle is the nucleus of a helium atom. It has a mass $m=6.64 \times 10^{-27}$kg and a charge $q = + 2e =  3.2 \times 10^{-19}$ C. Compare the force of the electric repulsion between two a-particles with the force of gravitational attraction between them.

physics magnetic fields and electromagnetism contact and non-contact forces comparing force in magnetic, electric and gravitational fields identifying forces

  1. $\displaystyle \dfrac{F _e}{F _g} = 3.1 \times 10^{32}$

  2. $\displaystyle \dfrac{F _e}{F _g} = 3.1 \times 10^{35}$

  3. $\displaystyle \dfrac{F _e}{F _g} = 3.1 \times 10^{38}$

  4. $\displaystyle \dfrac{F _e}{F _g} = 1 \times 10^{32}$

Show answer
Correct Option: B
Explanation:

Electrostatic repulsion = $F _e=\dfrac{1}{4 \pi \epsilon } \dfrac{q^2}{r^2 }=\dfrac{1}{4 \pi \epsilon } \dfrac{4e^2}{r^2 } $

gravitational attraction = $ F _g = G\dfrac{m^2}{r^2} $

$\dfrac{F _e}{F _g} = \dfrac{1}{4 \pi \epsilon } \dfrac{4 e^2}{G m^2} = 3.1 \times 10^{35} $

Electrostatic force and gravitational force differ in which respect?

physics magnetic fields and electromagnetism contact and non-contact forces comparing force in magnetic, electric and gravitational fields identifying forces

  1. Conservative force

  2. Central force

  3. Principle of superposition

  4. Dependence on the intervening medium

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Correct Option: D
Explanation:

In electrostatic force, the force of medium depend on charges while the force of medium due to gravity does not depend on masses. There is no concept of induced mass.

The ratio of electric force between two electrons to the gravitational force between them is of the order

physics magnetic fields and electromagnetism contact and non-contact forces comparing force in magnetic, electric and gravitational fields identifying forces

  1. 10$^{42}$

  2. 10$^{39}$

  3. 10$^{36}$

  4. 1

Show answer
Correct Option: A
Explanation:

The electric force between two electrons is $F _e=\dfrac{ke^2}{r^2}$

and gravitational force between two electrons is $F _g=\dfrac{Gm _e^2}{r^2}$
Thus, $\dfrac{F _e}{F _g}=\dfrac{ke^2}{Gm _e^2}=\dfrac{(9\times 10^9)(1.6\times 10^{-19})^2}{(6.67\times 10^{-11})(9.1\times 10^{-31})^2}=4.17\times 10^{42} \sim 10^{42}$