Question

An electron that is moving through a uniform magnetic field has a velocity v = (54 km/s)i + 48 km/s)j when it experiences a force F = (-3.800×10^-15 N)i + (4.275×10^-15 N)j due to the magnetic field.

a) If Bx = 0, calculate the magnitude of the magnetic field B. (Tesla = T)

b) What is By?

c) What is Bz?

Answer 1

Magnetic force is given by,

$F = q(v \times B)$

$\Rightarrow -3.8 \times 10^{-15} \ \hat{i} + 4.275 \times 10^{-15} \ \hat j = 1.62 \times 10^{-19} ((54000 \hat i + 48000 \hat j)\times B)$

Comparing each component and solving, since Bx = 0 ,

$B_y = 0$

since no k component in force and i x j = k, hence only z component in B,

$48000 B_z = -23456.79$

$\Rightarrow B_z = -0.488 \ T$

$|B| = 0.488 \ T$ Ans. of A part