A proton of velocity 3i^+2j^ms−1 enters a field of magnetic induction 2j^+3k^ tesla. The acceleration (in m/s2) produced in the proton is (charge to mass ratio of proton = 0.96 × 108 C kg-1)

F→=qV→×B→⇒a→=F→m=qmV→×B→=i^j^k^320023qm=0.96×1086i^−9j^+6k^=2.88×1082i^−3j^+2k^ m/s2

A proton of velocity 3i^+2j^ms−1 enters a field of magnetic induction 2j^+3k^ tesla. The acceleration (in m/s2) produced in the proton is (charge to mass ratio of proton = 0.96 × 108 C kg-1)

F→=qV→×B→⇒a→=F→m=qmV→×B→=i^j^k^320023qm=0.96×1086i^−9j^+6k^=2.88×1082i^−3j^+2k^ m/s2

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A proton of velocity $(3 \hat{i} + 2 \hat{j}) {ms}^{- 1}$ enters a field of magnetic induction $(2 \hat{j} + 3 \hat{k})$ tesla. The acceleration (in m/s²) produced in the proton is (charge to mass ratio of proton = $0.96 \times 10^{8} C k g^{- 1}$ )

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$2.8 \times 10^{8} (2 \hat{i} - 3 \hat{j})$

$2.88 \times 10^{8} (2 \hat{i} - 3 \hat{j} + 2 \hat{k})$

$2.8 \times 10^{8} (2 \hat{i} + 3 \hat{k})$

$2.88 \times 10^{8} (\hat{i} - 3 \hat{j} + 2 \hat{k})$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} \vec{F} = q (\vec{V} \times \vec{B}) \\ \Rightarrow \vec{a} = \frac{\vec{F}}{m} = \frac{q}{m} (\vec{V} \times \vec{B}) = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 0 \\ 0 & 2 & 3 \end{matrix}| \frac{q}{m} \end{array}$

$\begin{array}{l} = 0 .96 \times 10^{8} [6 \hat{i} - 9 \hat{j} + 6 \hat{k}] \\ = 2 .88 \times 10^{8} [2 \hat{i} - 3 \hat{j} + 2 \hat{k}] m / s^{2} \end{array}$