When a proton has a velocity V¯=2i^+3j^×106m/s, it experience a force F¯=−1.28×10−13k^N When its velocity is along the z – axis, it experiences a force along the x – axis. The magnetic field is

qp=1.6×10−19C =2i^+3j^×106 ms−1 F¯=−1.28×10−13k^N B¯=? When v is along z – axis, F¯ is along – x axis. B¯ must be in the y – z plane B¯=Byj^+Bzk^ Then F¯=qpv¯×B¯ ⇒−1.28×10−13k^=+1.6×10−19i^j^k^2300By0106 −1.28×10−131.6×10−19×106k^=i^0−j^0+k^2By −0.8k^=2Byk^ By=−0.4 B¯=Byj^=−0.4j^T ∴B¯=−0.4j^.T

When a proton has a velocity V¯=2i^+3j^×106m/s, it experience a force F¯=−1.28×10−13k^N When its velocity is along the z – axis, it experiences a force along the x – axis. The magnetic field is

qp=1.6×10−19C =2i^+3j^×106 ms−1 F¯=−1.28×10−13k^N B¯=? When v is along z – axis, F¯ is along – x axis. B¯ must be in the y – z plane B¯=Byj^+Bzk^ Then F¯=qpv¯×B¯ ⇒−1.28×10−13k^=+1.6×10−19i^j^k^2300By0106 −1.28×10−131.6×10−19×106k^=i^0−j^0+k^2By −0.8k^=2Byk^ By=−0.4 B¯=Byj^=−0.4j^T ∴B¯=−0.4j^.T

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When a proton has a velocity $\bar{V} = (2 \hat{i} + 3 \hat{j}) \times 10^{6} m/s$ , it experience a force $\bar{F} = - (1.28 \times 10^{- 13} \hat{k}) N$ When its velocity is along the z – axis, it experiences a force along the x – axis. The magnetic field is

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$\bar{B} = (- 0.4 \hat{j}) T$

$\bar{B} = (0.6 \hat{i}) T$

$\bar{B} = (8.0 \hat{j}) T$

$\bar{B} = (5 \hat{k}) T$

answer is A.

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Detailed Solution

$q_{p} = 1.6 \times 10^{- 19} C = (2 \hat{i} + 3 \hat{j}) \times 10^{6} {ms}^{- 1} \bar{F} = - (1.28 \times 10^{- 13} \hat{k}) N \bar{B} = ? W h e n v i s a l o n g z - a x i s, \bar{F} i s a l o n g - x a x i s . \bar{B} m u s t b e i n t h e y - z p l a n e \bar{B} = B_{y} \hat{j} + B_{z} \hat{k} T h e n \bar{F} = q_{p} (\bar{v} \times \bar{B}) \Rightarrow - 1.28 \times 10^{- 13} \hat{k} = + 1.6 \times 10^{- 19} |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 0 \\ 0 & B_{y} & 0 \end{matrix}| 10^{6} \frac{- 1.28 \times 10^{- 13}}{1.6 \times 10^{- 19} \times 10^{6}} \hat{k} = \hat{i} (0) - \hat{j} (0) + \hat{k} (2 B_{y}) - 0.8 \hat{k} = 2 B_{y} \hat{k} B_{y} = - 0.4 \bar{B} = B_{y} \hat{j} = (- 0.4 \hat{j}) T ∴ \bar{B} = - (0.4) \hat{j} . T$