A proton moves perpendicular to a uniform magnetic field B at 1.20 107 m/s and experiences an acceleration of 2.00 1013 m/s2 in the +x direction when its velocity is in the +z direction. Determine the magnitude and direction of the field.

Question

A proton moves perpendicular to a uniform magnetic field B at 1.20  107 m/s and experiences an acceleration of 2.00  1013 m/s2 in the +x direction when its velocity is in the +z direction. Determine the magnitude and direction of the field.

Elena · Answer

ma=qvBsinα sinα=1 q=e ma=evB B=ma/ev = =1.67•10⁻²⁷•2 •10¹³/1.6•10⁻¹⁹•1.2•10⁷= =0.0173 T F=q[v,B] => B is in –y direction