Question

An electron that is moving through a uniform magnetic field has a velocity v = (52 km/s)i + 40 km/s)j when it experiences a force F = (-3.600×10^-15 N)i + (4.680×10^-15 N)jdue to the magnetic field. If Bx = 0, calculate the magnitude of the magnetic field B. (Tesla = T). What is By? What is Bz?

Answer 1

V =( 52 i + 40 j )*10³ m/s
F =( -3.6 i + 4.68 j )*10^-15 N
We know that
Magnetic force on the charge is given by
F = q [V x B]
B = B_X i + B_Y j + B_z k
Where B_x = 0 , therefore
B = 0 i + B_Y j + B_Z k
where q is charge on electron = 1.6*10^-19
F = (1.6*10^-19)[( 52*10³ i + 40*10³ j ) x (0 i + B_Y j + B_Z k ) ]
On solving we get
( -3.6 i + 4.68 j )*10^-15 = 1.6*10^-19*10³[40 B_Z i - 52B_Z j + 52B_Y k ]
On compairing
B_Y = 0
B_Z = -0.5625 T
Now the magnitude of the B
B = 0.5625 T