An electron has velocity V¯=(2×106m/s)i^+(3×106m/s)j^Magnetic field present in the region is B¯=0.03Ti^−0.15Tj^.The force on the electron is

v¯=2×106i^+3×106j^m/s B¯=0.03i^−0.15j^T q=−1.6×10−19C F¯=? F¯=qv¯×B¯ F¯=−1.6×10−19i^j^k^2×1063×10600.03−0.150 =−1.6×10−192×106−0.15−3×106×0.03k^ =0.48+0.144k^×10−13 =0.624×10−13k^ =6.24×10−14k^

An electron has velocity V¯=(2×106m/s)i^+(3×106m/s)j^Magnetic field present in the region is B¯=0.03Ti^−0.15Tj^.The force on the electron is

v¯=2×106i^+3×106j^m/s B¯=0.03i^−0.15j^T q=−1.6×10−19C F¯=? F¯=qv¯×B¯ F¯=−1.6×10−19i^j^k^2×1063×10600.03−0.150 =−1.6×10−192×106−0.15−3×106×0.03k^ =0.48+0.144k^×10−13 =0.624×10−13k^ =6.24×10−14k^

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

An electron has velocity $\bar{V} = (2 \times 10^{6} m/s) \hat{i} + (3 \times 10^{6} m/s) \hat{j}$ Magnetic field present in the region is $\bar{B} = 0.03 T \hat{i} - 0.15 T \hat{j}$ .The force on the electron is

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

$(6.24 \times 10^{- 14} N) \hat{k}$

$(3.56 \times 10^{- 14} N) \hat{k}$

$(8 \times 10^{- 14} N) \hat{k}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\bar{v} = (2 \times 10^{6} \hat{i} + 3 \times 10^{6} \hat{j}) m/s \bar{B} = (0.03 \hat{i} - 0.15 \hat{j}) T q = - {1.6×10}^{- 19} C \bar{F} = ? \bar{F} = q (\bar{v} \times \bar{B}) \bar{F} = - 1.6 \times 10^{- 19} |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 \times 10^{6} & 3 \times 10^{6} & 0 \\ 0.03 & - 0.15 & 0 \end{matrix}| = - 1.6 \times 10^{- 19} [2 \times 10^{6} (- 0.15) - 3 \times 10^{6} \times 0.03] \hat{k} = (0.48 + 0.144) \hat{k} \times 10^{- 13} = 0.624 \times 10^{- 13} \hat{k} = (6.24 \times 10^{- 14}) \hat{k}$