Complete Question
An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an electron is 1.60218 × 10−19 C and its mass is 9.10939 × 10−31 kg. How strong a magnetic field must be experienced by the electron if its path is a circle of radius 5.4 cm?
Answer:
The magnetic field strength is 
Explanation:
The work done by the potential difference on the electron is related to the kinetic energy of the electron by this mathematical expression
Making v the subject
![v = \sqrt{[\frac{2 \Delta V * q }{m}] }](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5B%5Cfrac%7B2%20%5CDelta%20V%20%2A%20q%20%7D%7Bm%7D%5D%20%7D)
Where m is the mass of electron
v is the velocity of electron
q charge on electron
is the potential difference
Substituting values
![v = \sqrt{\frac{2 * 5.9 *10^3 * 1.60218*10^{-19} }{9.10939 *10^{-31]} }](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B%5Cfrac%7B2%20%2A%205.9%20%2A10%5E3%20%2A%201.60218%2A10%5E%7B-19%7D%20%7D%7B9.10939%20%2A10%5E%7B-31%5D%7D%20%7D)
f
For the electron to move in a circular path the magnetic force[
] must be equal to the centripetal force[
] and this is mathematically represented as
making B the subject
r is the radius with a value = 5.4cm = 
Substituting values
Answer:
C) upward
Explanation:
The problem can be solved by using the right-hand rule.
First of all, we notice at the location of the negatively charged particle (above the wire), the magnetic field produced by the wire points out of the page (because the current is to the right, so by using the right hand, putting the thumb to the right (as the current) and wrapping the other fingers around it, we see that the direction of the field above the wire is out of the page).
Now we can apply the right hand rule to the charged particle:
- index finger: velocity of the particle, to the right
- middle finger: direction of the magnetic field, out of the page
- thumb: direction of the force, downward --> however, the charge is negative, so we must reverse the direction --> upward
Therefore, the direction of the magnetic force is upward.
Answer:
Explanation:
a ) It is given that bomb was at rest initially , so , its momentum before the explosion was zero.
b ) We shall apply law of conservation of momentum along x and y direction separately because no external force acts on the bomb.
If v be the velocity of the third part along a direction making angle θ
with x axis ,
x component of v = vcosθ
So momentum along x axis after explosion of third part = mv cosθ
= 10 v cosθ
Momentum along x of first part = - 5 x 42 m/s
momentum of second part along x direction =0
total momentum along x direction before explosion = total momentum along x direction after explosion
0 = - 5 x 42 + 10 v cosθ
v cosθ = 21
Similarly
total momentum along y direction before explosion = total momentum along y direction after explosion
0 = - 5 x 38 + 10 v sinθ
v sinθ= 21
squaring and and then adding the above equation
v² cos²θ +v² sin²θ = 21² +19²
v² = 441 + 361
v = 28.31 m/s
Tanθ = 21 / 19
θ = 48°
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