The magnetic field and electric field in a region in space are \(\vec

1.	The magnetic field and electric field in a region in space are \(\vec B\) = B\(\hat i\) and \(\vec E\) = E\(\hat i\).A particle of charge q moves into the region with velocity \(\vec v\) = v\(\hat j\). Find the magnitude and direction of the Lorentz force on the charged particle if q = 1 C,B = 1 T,E = 3 V/m and v = 4 m/s,
Answer» Data : q = 1 C, B = 1 T, E = 3 V/m, v = 4 m/s The Lorentz force on the charged particle is \(\vec F_m\) = q (\(\vec E\) + \(\vec v\) x \(\vec B\)) = q[E\(\hat i\) + (v\(\hat j\) x B\(\hat i\))] = q [E\(\hat i\)+ vB(-\(\hat k\))] = (1) [3\(\hat i\) - 4\(\hat k\)] = 3\(\hat i\) - 4\(\hat k\) N Therefore, \(\vec F_m\) is in the y = 0 plane (i.e, in the x - z plane). \(\therefore\) \(F_m\) = \(\sqrt{3^2+4^2}\) = 5 N If \(\vec F_m\) makes an angle \(\theta\) with the + x-axis, \(\theta\) = tan^-1 \(\left(-\cfrac43 \right)\) = - 53\(^\circ\)8' Therefore, \(\vec F_m\) is at an angle of 53\(^\circ\)8' to the x-axis towards the - z- axis.

The magnetic field and electric field in a region in space are \(\vec B\) = B\(\hat i\) and \(\vec E\) = E\(\hat i\).A particle of charge q moves into the region with velocity \(\vec v\) = v\(\hat j\). Find the magnitude and direction of the Lorentz force on the charged particle if q = 1 C,B = 1 T,E = 3 V/m and v = 4 m/s,

Answer»

Data : q = 1 C,

B = 1 T, E = 3 V/m,

v = 4 m/s

The Lorentz force on the charged particle is

\(\vec F_m\) = q (\(\vec E\) + \(\vec v\) x \(\vec B\))

= q[E\(\hat i\) + (v\(\hat j\) x B\(\hat i\))]

= q [E\(\hat i\)+ vB(-\(\hat k\))]

= (1) [3\(\hat i\) - 4\(\hat k\)]

= 3\(\hat i\) - 4\(\hat k\) N

Therefore, \(\vec F_m\) is in the y = 0 plane (i.e, in the x - z plane).

\(\therefore\) \(F_m\) = \(\sqrt{3^2+4^2}\) = 5 N

If \(\vec F_m\) makes an angle \(\theta\) with the + x-axis,

\(\theta\) = tan^-1 \(\left(-\cfrac43 \right)\)

= - 53\(^\circ\)8'

Therefore, \(\vec F_m\) is at an angle of 53\(^\circ\)8' to the x-axis towards the - z- axis.