Physics, 13.08.2019 05:20 srshelton

Aproton moves perpendicular to a uniform magnetic field b with arrow at a speed of 2.20 107 m/s and experiences an acceleration of 1.90 1013 m/s2 in the positive x-direction when its velocity is in the positive z-direction. determine the magnitude and direction of the field.

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Answer from: kisses1131953

The magnitude and direction of the magnetic field is 0.009014 T in the negative y direction.

Explanation:

Given that,

Speed $v = 2.20\times10^7\ m/s$

Acceleration $a=1.90\times10^{13}\ m/s^2$

We need to calculate the magnetic field

Using formula of magnetic field

F=qvB ....(I)

Using newton's second law

F= ma ....(II)

From equation (I) and (II)

ma=qvB

Put the value into the formula

$1.90\times10^{13}\times1.67\times10^{-27}=1.6\times10^{-19}\times2.20\times10^{7}\timesB$

$3.173\times10^{-14}=1.6\times10^{-19}\times2.20\times10^{7}\timesB$

$B=\dfrac{3.173\times10^{-14}}{1.6\times10^{-19}\times2.20\times10^{7}}$

$B=0.009014\ T$

We need to calculate the direction of the field

Using the right hand rule, point the right hand fingers along the velocity which is in the positive z direction.

Now, if we curl the fingers along the direction of magnetic field that is in the negative y direction, then the thumb will point in the positive x direction.

Hence, The magnitude and direction of the magnetic field is 0.009014 T in the negative y direction.

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Aproton moves perpendicular to a uniform magnetic field b with arrow at a speed of 2.20 107 m/s and...

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