Question

A particle that has an 7.3-μC charge moves with a velocity of magnitude 4 × 105 m/s along the +x axis. It experiences no magnetic force, although there is a magnetic field present. The maximum possible magnetic force that the charge with the given speed could experience has a magnitude of 0.320 N. Find the magnitude and direction of the magnetic field. Note that there are two possible answers for the direction of the field.

Answer 1

If the particle's velocity were perpendicular to the magnetic field, the magnetic force would be given by:

F = qvB

F is the magnetic force, q is the charge, v is the velocity, and B is the magnetic field strength.

Given values:

F = 0.320N

q = 7.3×10⁻⁶C

v = 4×10⁵m/s

Plug in the values and solve for B:

0.320 = (7.3×10⁻⁶)(4×10⁵)B

B = 0.110T

The magnetic force depends on the cross product between the particle's velocity vector v and the magnetic field vector B. If v and B are parallel then the cross product is 0. v points in the +x direction, therefore B must point in either the +x or -x direction.

Answer 2

The magnitude of the magnetic field is 0.11 T and the direction of the magnetic field must be either y-direction or z-direction since the velocity is in x-direction.

The given parameters;

• charge of the particle, q = 7.3 μC
• velocity of the particle, v = 4 x 10⁵ m/s
• magnetic force on the particle, F = 0.32 N

The magnitude of the magnetic field is calculated as follows;

F = q(v x B)

$B = \frac{F}{qv} \\\\B = \frac{0.32}{7.3 \times 10^{-6} \times 4\times 10^5} \\\\B = 0.11 \ T$

Since, the magnetic force experienced by the charge is maximum, then the magnetic field must be perpendicular to the velocity of the charge.

Thus, the direction of the magnetic field must be either y-direction or z-direction since the velocity is in x-direction.

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