Question

At time t=0 a positively charged particle of mass m=3.57 g and charge q=9.12 µC is injected into the region of the uniform magnetic B=B k and electric E=−E k fields with the initial velocity v=v0 i. The magnitudes of the fields: B=0.18 T, E=278 V/m, and the initial speed v0=2.1 m/s are given. Find at what time t, the particle's speed would become equal to v(t)=3.78·v0:

Answer 1

10.78 s

Step-by-step explanation:

The force on the charge is computed by using the equation:

`F^(\to)= qE^(\to) +q (v^(\to) + B^(\to)) \\ \\ F^(\to) = (9.12 * 10^(-6)) *278 (-\hat k) +9.12 *10^(-6) *2.1 *0.18 (\hat i * \hat k) \\ \\ F^(\to) = -2.535 *10^(-3) \hat k -3.447*10^(-6) \hat j`

F = ma

∴

`a ^(\to)= (F^(\to))/(m)`

`a ^(\to)= (-1)/(3.57* 10^(-3))(2.535*10^(-3)\hat k + 3.447*10^(-6) \hat j)`

`a ^(\to)=-0.710 \hat k -9.656*10^(-4) \hat j`

At time t(sec; the partiCle velocity becomes
`v(t) = 3.78 v_o`

The velocity of the charge after the time t(sec) is expressed by using the formula:

`v^(\to)= v_(o \ \hat i) + a^(\to )t \\ \\ \implies (2.1)\hat i -0.710 t \hat k -9.656 * 10^(-4) t \hat j = 3.78 v_o \\ \\ \implies (2.1)^2 +(0.710\ t)^2+ (9.656 *10^(-4)t )^2 = (3.78 *2.1^2 \\ \\ \implies 4.41 +0.5041 t^2 +9.324*10^(-7) t^2 = 63.012 \\ \\ \implies 4.41 +0.5041 t^2 = 63.012\\ \\ 0.5041t^2 = 63.012-4.41 \\ \\ t^2 = (58.602)/(0.5041) \\ \\ t^2 = 116.25 \\ \\ t = √(116.25) \\ \\ \mathbf{t = 10.78 \ s}`