A proton moves with a velocity of 5x106 m/s in the +y direction as shown. What i

Question

A proton moves with a velocity of 5x106 m/s in the +y direction as shown. What is the force (magnitude and direction) on the proton if:

a. A magnetic field of 3T is applied in the +x direction.

b. A magnetic field of 3T is applied in the +z direction.

c. A magnetic field of 2.12Ti + 2.12Tj is applied.

A proton moves with a velocity of 5x106 m/s in the +y direction as shown. What is the force (magnitude and direction) on the proton if: A magnetic field of 3T is applied in the +x direction. A magnetic field of 3T is applied in the +z direction. A magnetic field of 2.12Ti + 2.12Tj is applied.

Explanation / Answer

a). F = q.(vxB) = qvB*sin(theta) direction is given by cross product of velocity vector and B vector. F = 1.6*10^-19*5*10^6*3*sin90 = 2.4*10^-12 N (in -z direction) b). F = 1.6*10^-19*5*10^6*3*sin90 = 2.4*10^-12 N (in +x direction) c). F1 (due to B = 2.12T i)= 1.6*10^-19*5*10^6*2.12*sin90 = 1.696*10^-12 N (in -z direction) the angle between v and 2.12T j is 0 degrees=> sin0 = 0 so, F2 (due to B = 2.12T j)= 1.6*10^-19*5*10^6*2.12*sin0 = 0 N total F = F1+F2 = 1.696*10^-12 N (in -z direction)

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