Question

An electron moves at 2.70×10⁶ m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.10×10⁻² T What is the largest possible magnitude of the acceleration of the electron due to the magnetic field IN M/S^2?)

Answer 1

Since , force experienced by electron in magnetic field is given by

F = q v B sin

Also, F = m a

Where, q is charge of electron = 1.6 × 10^-19 C

v is velocity of electron = 2.70 × 10⁶ m/s

B is magnitude of magnetic field = 7.10 × 10^-2 T

m is the mass of electron = 9.1 × 10 ^-31 kg

For greatest possible magnitude of acceleration , = 90°

So , a = q v B / m = 1.6 × 10^-19 × 2.70 × 10⁶ × 7.10 × 10^-2 / 9.1 × 10^-31   = 3.37 × 10¹⁶ m/s²