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

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

the results is a decrease in air temperature with an increase of altitude.

explanation:

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft