A proton moving in the XY plane with a velocity of 6×10^7m/s makes an angle of 30° with a uniform magnetic field of induction B=1.5T acting the positive direction of Y-axis. Find the force acting upon it.

Question

Rohit · Answer

Use formula F = qvB sinx. Here x= 30°. Substitute the values given in the question. U will gee the answer. Use q = 1.6×10^-19 C. Hoping that it will help u. Regards Rohit.

Tiamiyu · Answer

An electron is moving in the xy- plane with a velocity of 3.0×10^5 m/s directed at an angle 20° above the positive x- axis .impressed on the electron is a 7.5mT find the magnitude and direction of the force on the electron if the direction of the field is along (I)positive x axis (ii) negative y axis (iii) the positive z axis

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A proton moving in the XY plane with a velocity of 6×10^7m/s makes an angle of 30° with a uniform magnetic field of induction B=1.5T acting the positive direction of Y-axis. Find the force acting upon it.

A proton moving in the XY plane with a velocity of 6×10^7m/s makes an angle of 30° with a uniform magnetic field of induction B=1.5T acting the positive direction of Y-axis. Find the force acting upon it.

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