If a proton of mass 1.67262 X 10 -27 Kg and velocity= 3.81 X 10 7 m/s continues

Question

If a proton of mass 1.67262 X 10-27 Kg and velocity= 3.81 X 107 m/s continues to move in a directionthat is consistently perpendicular to a magnetic field = 0.434 T,what is the radius of curvature of its path?? Is this question asking for the radius of the circle thatwould form from the path of the proton??? If a proton of mass 1.67262 X 10-27 Kg and velocity= 3.81 X 107 m/s continues to move in a directionthat is consistently perpendicular to a magnetic field = 0.434 T,what is the radius of curvature of its path?? Is this question asking for the radius of the circle thatwould form from the path of the proton???

Explanation / Answer

   radius of curvature of pathtaken   r   =   m * v/ B * q    r   =   1.67262 *10-27 * 3.81 * 107 / 0.434 * 1.6 *10-19    r   =   9.17725 *10-1   =   0.917725   m    r   =   9.17725 *10-1   =   0.917725   m

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