Question

A particle with charge -5.60 nC is moving in a uniform magnetic field →B=−(1.25T) k The magnetic force on the particle is measured to be →F=−(3.40×10−7N) ˆı +(7.40×10−7N) ˆȷ(a) Calculate all the components of the velocity of the particle that you can from this information. (b) Are there components of the velocity that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product →v ⋅ →F . What is the angle between →v and →F

Answer 1

Answer:

Explanation:

Force = q ( v x B)

- 5.6 x 10⁻⁹ (v x - 1.25 k )

- 3.4x 10⁻⁷i + 7.4 x 10⁻⁷j

Let v = ai+bj +ck

Force = - 5.6 x 10⁻⁹ [(ai+bj +ck) x - 1.25 k )]

= - 5.6 x 10⁻⁹ ( 1.25aj - 1.25bi )

= - 7 a j + 7 b i

( 7bi - 7aj ) x 10⁻⁹

Comparing with given force

7b x 10⁻⁹ b = - 3.4 x 10⁻⁷

b = - 48.57

- 7 a x 10⁻⁹ = 7.4 x 10⁻⁷

a = - 105.7

velocity

= -105.7 i - 48.57 j + ck

b ) Component along k can not be obtained .

c ) v . F = ( -105.7 i - 48.57 j + ck ) . −(3.40×10−7N) ˆı +(7.40×10−7N) ˆȷ

= 105.7 x 3.4 x 10⁻⁷ - 48.57 x 7.4 x 10⁻⁷

= 359.38 x 10⁻⁷ - 359.38 x 10⁻⁷

=0

angle between v and F = 90 degree