Practice: Magnets And Charges In Motion

The magnitude of the force on the charge is given by the equation $$F=qvB\sin \theta,$$ where $q$ is the charge, $v$ is the charge's speed, $B$ is the magnitude of the magnetic field, and $\theta$ is the angle the magnetic field makes with the charge's velocity. Plugging in the values given, we get $$F=\left(2 \times 10^{-9} \ \mathrm{C}\right) \left( 10 \ \mathrm{\frac{m}{s}} \right) \left(10^{-5} \ \mathrm{T} \right)\sin 30^{\circ} $$ $$=\boxed{1 \times 10^{-13} \ \mathrm{N}} = 100 \ \mathrm{fN}. \ \ \mathrm{(100 \ femtonewtons)}$$

Insert to the equation : F = Q x V x B x sin tetha

• F = magnitude of the force (Newton / N)
• Q = charge (Coulomb / C)
• V = speed (Meter per Second / m/s)
• B = Magnetic Field (Tesla / T)

F = Q x V x B x sin tetha

F = 2^{-9} x10 x 10^{-5} x sin 30

F = 10^{-13}

Lorentz Force Theory on moving particles

$F = q \times V \times B \times sin \theta$

$F = 2 \times 10^{-9} \times 10 \times 10^{-5} \times 1/2$

$F = 1 \times 10^{-13}$

f=Bqvsin (thita) here, B=10^-5 T,v=10mps,sin(thita)=1/2 so f= 1E-13.

F=QVB Sin(angle subtended)

force acted on the charge is q(B X v) =q cross product of vectors(B,v) =q B v sin(angle) =2E-9 1E-5 10*0.5 =1E-13

F=qvbsin(theta) q=2E-9 v=10m/s b=E-5 sin 30=.5 f={2 10^(-9) 10 10^(-5) 0.5}=1E-13

We Know that

Force(F)=Bqvsin(theta)

So, F=1E-5* 2E-9 *10 *Sin30

      F=1E-13Newtons....

F= q(v *B)

Aplicação direta da fórmula da força magnética na carga: F=q.V.B.senΘ, sendo F a força, q a carga, V a velocidade e B o campo magnético. F=2E-9 x 10 x 1E-5/2 ::F=1E-13

F=q(VxB) F=qVBSin(a) Given, q=2x10^-9 V=10 B=10^-5 a=30 therefore, F=2x10^-9x10x10^-5xSin(30) F=1x10^-13 N

We know, from the formula of the Lorentz Force that :

F = Q (v $\times B)$ , where Q is the charge, v is the velocity and B is the Magnetic Flux Density and the $\times$ is the Vector Cross Product .

Also, a $\times$ b = absin $\theta$

Now, putting the values in we get: F = $2*10^{-9}*(10^{-5})*10*sin(30) = \boxed{1*10^{-13}}$ N

$F=|q| \times |B| \times |v| \times \sin \alpha$ , where $\alpha$ is the angle formed by $B$ and $v$ .

If we do this equation, we'll get $\boxed{10^{-13}}$ .

It is well-known that $F = qvB \sin \theta$ . Using $q = 2 \cdot 10^{-9} \, \text{C}$ , $v = \text {10 m/s}$ , $B = 10^{-5} \, \text {T}$ , and $\theta = 30^\circ$ , we get $F = \boxed {1 \cdot 10^{-13} \, \text{N}}$ .

Since We know that

F=qvBsin(angle)

Where

B= Magnetic field

v=velocity of charge

q=magnitude of charge

By putting values we get the answer

The force is $qvBsin\theta = (2 \times 10^{-9})(10)(10^{-5})\sin{30^\circ} = 1\times 10^{-13} \text{N}$

$\text{Force}, F=qvB\sin{\theta}$ , where $q$ is the charge, $v$ is the velocity, $B$ is the the magnetic field and $\theta$ is the angle between the velocity and the magnetic field.

Hence $F=2\times 10^{-9}\times 10\times 10^{-5}\times \sin{30^{\circ}}=1\times 10^{-13}$ .