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

Answer:

10.78 s

Explanation:

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

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

F = ma

∴

$a ^{\to}= \dfrac{F^{\to}}{m}$

$a ^{\to}= \dfrac{-1}{3.57\times 10^{-3}}(2.535*10^{-3}\hat k + 3.447*10^{-6} \hat j)$

$a ^{\to}=-0.710 \hat k -9.656*10^{-4} \hat j$

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:

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