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jek_recluse [69]
3 years ago
5

An electron moving with a velocity = 5.0 × 107 m/s enters a region of space where perpendicular electric and a magnetic fields a

re present. The electric field is = - 104 . What magnetic field will allow the electron to go through, undeflected?
Physics
1 answer:
Cloud [144]3 years ago
3 0

Answer:

magnetic field will allow the electron to go through 2 x 10^{4}  T k

Explanation:

Given data in question

velocity = 5.0 × 10^{7}

electric filed = 10^{4}

To find out

what magnetic field will allow the electron to go through, undeflected

solution

we know if electron move without deflection i.e. net force is zero on electron and we can say both electric and magnetic force equal in magnitude and opposite in directions

so we can also say

F(net) = Fe + Fb i.e. = 0

q V B +  q E = 0

q will be cancel out

10^{4}j + 5e + 7i × B =  0

B = 2 x 10^{4}  T k

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6 0
2 years ago
he atomic radii of Li and O2- ions are 0.068 and 0.140 nm, respectively. (a) Calculate the force of attraction in newtons betwee
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Answer:

a

 F =  -1.07 *10^{-8} \  N

b

F_r  =  1.07 *10^{-8} \  N

Explanation:

Generally the force of attraction between this two irons is mathematically represented as

F =  \frac{k *  [Q_{Li}  ] * [Q_{O}  ]  }{ r^2}

Here k is known as the proportionality constant with value k = 2.31  *  10^ {-28} J \cdot m

substituting -2 for Q_{O} i.e the charge on oxygen , +1 for Q_{Li} i.e the charge on Lithium and [0.140 + 0.068 ] nm= 0.208 nm =  0.208*10^{-9} for r

So

F =  \frac{ 2.31  *  10^ {-28}*  +1   * -2   }{ ( 0.208*10^{-9} )^2   }

F =  -1.07 *10^{-8} \  N

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So Force of repulsionn is

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