A proton of velocity (3hat i + 2hat j) , ms^{ | vedclass

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(B) Given velocity $\vec{v} = (3\hat i + 2\hat j) \, m/s$ and magnetic field $\vec{B} = (2\hat j + 3\hat k) \, T$.The magnetic force on a charged part

Vedclass · Accepted Answer

$2.88 	imes 10^8 (2\hat i - 3\hat j + 2\hat k) \, m/s^2$

Vedclass · Answer

(B) Given velocity $\vec{v} = (3\hat i + 2\hat j) \, m/s$ and magnetic field $\vec{B} = (2\hat j + 3\hat k) \, T$.The magnetic force on a charged particle is given by $\vec{F} = q(\vec{v} 	imes \vec{B})$.$\vec{v} 	imes \vec{B} = (3\hat i + 2\hat j) 	imes (2\hat j + 3\hat k) = 3\hat i 	imes 2\hat j + 3\hat i 	imes 3\hat k + 2\hat j 	imes 2\hat j + 2\hat j 	imes 3\hat k$Using cross product rules $(\hat i 	imes \hat j = \hat k, \hat i 	imes \hat k = -\hat j, \hat j 	imes \hat j = 0, \hat j 	imes \hat k = \hat i)$:$\vec{v} 	imes \vec{B} = 6\hat k - 9\hat j + 0 + 6\hat i = (6\hat i - 9\hat j + 6\hat k) = 3(2\hat i - 3\hat j + 2\hat k)$.Acceleration $\vec{a} = \frac{\vec{F}}{m} = \frac{q}{m}(\vec{v} 	imes \vec{B})$.Given $\frac{q}{m} = 0.96 	imes 10^8 \, C/kg$.$\vec{a} = (0.96 	imes 10^8) 	imes 3(2\hat i - 3\hat j + 2\hat k) = 2.88 	imes 10^8 (2\hat i - 3\hat j + 2\hat k) \, m/s^2$.

$A$ proton of velocity $(3\hat i + 2\hat j) \, ms^{-1}$ enters a magnetic field of $(2\hat j + 3\hat k) \, T$. The acceleration produced in the proton is (charge to mass ratio of proton $= 0.96 \times 10^8 \, C/kg$)

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