A particle of charge\(q > 0\)is moving at speed\(v\)in the\( + z\)-direction through a region of uniform magnetic field\(\vec B\). The magnetic force on the particle is\(\vec F = {F_0}\left( {3\hat i + 4\hat j} \right)\), where\({F_0}\)is a positive constant. (a) Determine the components\({B_x}\),\({B_y}\), and\({B_z}\), or at least as many of the three components as is possible from the information given. (b) If it is given in addition that the magnetic field has magnitude\(6{F_0}/qv\), determine as much as you can about the remaining components of\(\vec B\).

The term magnetic field may be defined as the area around the magnet behaves like a magnet.

The magnetic force of the particle is,\(\vec F = {F_0}\left( {3\hat i + 4\hat j} \right)\).

The magnetic field is, \(B = \frac{{6{F_0}}}{{qv}}\).

The force can be calculated as

\(\vec F = q\vec v \times \vec B\)

Put all the given values

\(\begin{aligned}\vec F = q\left( {v\hat k} \right) \times \left( {{B_x}\hat i + {B_y}\hat j + {B_z}\hat k} \right)\\\vec F = - qv{B_y}\hat i + qv{B_x}\hat j\\{F_0}\left( {3\hat i + 4\hat j} \right) = - qv{B_y}\hat i + qv{B_x}\hat j\end{aligned}\)

Compare both sides get

\(\begin{aligned}3{F_0} = - qv{B_y}\\{B_y} = - \frac{{3{F_0}}}{{qv}}\\4{F_0} = qv{B_x}\\{B_x} = \frac{{4{F_0}}}{{qv}}\end{aligned}\)

Hence, the component of \({B_y} = - \frac{{3{F_0}}}{{qv}},{B_x} = \frac{{4{F_0}}}{{qv}}\).

The magnitude of the magnetic field is calculated as,

\(\begin{aligned}{c}B = \frac{{6{F_0}}}{{qv}}\\\sqrt {B_x^2 + B_y^2 + B_z^2} = \frac{{6{F_0}}}{{qv}}\end{aligned}\)

Substitute all the value in the above equation.

\(\begin{aligned}\sqrt {{{\left( {\frac{{4{F_0}}}{{qv}}} \right)}^2} + {{\left( { - \frac{{3{F_0}}}{{qv}}} \right)}^2} + B_z^2} = \frac{{6{F_0}}}{{qv}}\\6 = \sqrt {9 + 16 + {{\left( {\frac{{qv}}{{{F_0}}}} \right)}^2}B_z^2} \\36 = 25 + {\left( {\frac{{qv}}{{{F_0}}}} \right)^2}B_z^2\end{aligned}\)

\(\begin{aligned}11 = {\left( {\frac{{qv}}{{{F_0}}}} \right)^2}B_z^2\\{B_z} = \pm \frac{{\sqrt {11} {F_0}}}{{qv}}\end{aligned}\)

Hence, the remaining component of\(\vec B\)is \( \pm \frac{{\sqrt {11} {F_0}}}{{qv}}\).