Physics
08.08.2021 17:22
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A beam of protons is directed in a straight line along the positive zz ‑direction

A beam of protons is directed in a straight line along the positive zz ‑direction through a region of space in which there are crossed electric and magnetic fields. If the electric field magnitude is E=450E=450 V/m in the negative yy ‑direction and the protons move at a constant speed of v=7.9×105v=7.9×105 m/s, what must the direction and magnitude of the magnetic field be in order for the beam of protons to continue undeflected along its straight-line trajectory? Select the direction of the magnetic field BB .
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alopez117
alopez117
4,7(63 marks)

The magnitude is B = \frac{450}{7.9* 10^5}

The direction is  the positive x axis

Explanation:

From the question we are told that

   The  electric field is  E = 450 V/m in the negative y ‑direction

   The speed of the proton is  v= 7.9* 10^5\  m/s in the positive z direction

Generally the overall force acting on  the proton is mathematical represented as

              F_E =  q(\vec E + \vec v  * \vec B)

Now for the beam of protons to continue un-deflected along its straight-line trajectory then  F_E =0

So

          0  =  q( E (-y) + v(z)  * \vec B)

=>      E\^y = v \^ z  * \vec B

Generally from unit vector cross product vector multiplication

         \^ z  \ *  \  \^ x  =  \^  y

So the direction of  B (magnetic field must be in the positive x -axis )

So

       E\^y = v \^ z  *  B\^ x

=>     E\^y = vB ( \^ z  *  \^ x)

=>     E\^y = vB ( \^y)        

=>      E = vB  

=>      B = \frac{450}{7.9* 10^5}

=>      B =  0.0005696 \ T  

   

   

Whitehouse9
Whitehouse9
4,6(88 marks)
Newton's third law explains the generation of thrust by a rocket engine.

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