Answer:
Explanation:
We shall show all given data in vector form and calculate the direction of force with the help of following formula
force F = q ( v x B )
q is charge , v is velocity and B is magnetic field.
Given B = - Bk ( i is right , j is upwards and k is straight up the page )
v = v j
F = q ( vj x - Bk )
= -Bqvi
The direction is towards left .
a ) If velocity is down
v = - v j
F = q ( - vj x - bk )
= qvB i
Direction is right .
b ) v = v i
F = q ( vi x - Bk )
= qvB j
force is upwards
c ) v = - vi
F = q ( -vi x - Bk )
= -qvBj
force is downwards
d ) v = - v k
F = q( - vk x -Bk )
= 0
No force will be created
e ) v = v k
F = q( vk x -Bk )
= 0
No force will be created
Answer:
F = 63N
Explanation:
M= 1.5kg , t= 2s, r = (2t + 10)m and
Θ = (1.5t² - 6t).
magnitude of the resultant force acting on 1.5kg = ?
Force acting on the mass =
∑Fr =MAr
Fr = m(∇r² - rθ²) ..........equation (i)
∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)
The horizontal path is defined as
r = (2t + 10)
dr/dt = 2, d²r/dt² = 0
Angle Θ is defined by
θ = (1.5t² - 6t)
dθ/dt = 3t, d²θ/dt² = 3
at t = 2
r = (2t + 10) = (2*(2) +10) = 14
but dr/dt = 2m/s and d²r/dt² = 0m/s
θ = (1.5(2)² - 6(2) ) = -6rads
dθ/dt =3(2) - 6 = 0rads
d²θ/dt = 3rad/s²
substituting equation i into equation ii,
Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)
∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]
∑F = 1.5(14*3+0) = 63N
F = √(Fr² +FΘ²) = √(0² + 63²) = 63N