Questions

An experimenter’s diary reads as follows: “A charged particle is projected in a magnetic field of $(7 \cdot 0 \hat{i} - 3 \cdot 0 \hat{j}) \times 10^{- 3}$ T. The acceleration of the particle is found to be $(x \hat{i} + 7 \cdot 0 \hat{j}) \times 10^{- 6} {ms}^{- 2}$ ." The number to the left of $\hat{i}$ in the last expression was not readable. Find the value of x ?

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detailed solution

Correct option is C

$\begin{array}{l} \vec{B} = (7 .0 \hat{i} - 3 .0 \hat{j}) \times 10^{- 3} T \\ \vec{a} = Acceleration = (- \hat{i} + 7 \hat{j}) \times 10^{- 6} m / s^{2} \end{array}$
Let the gap be x .

Since $\vec{B} and \vec{a}$ are always perpendicular
$\begin{array}{l} \vec{B} \times \vec{a} = 0 \\ \Rightarrow (7 \times x \times 10^{- 3} \times 10^{- 6} - 3 \times 10^{- 3} \times 7 \times 10^{- 6}) = 0 \\ \Rightarrow 7 x - 21 = 0 \Rightarrow x = 3 \end{array}$

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detailed solution

Correct answer is 3

$\begin{array}{l} \vec{B} = (7 .0 \hat{i} - 3 .0 \hat{j}) \times 10^{- 3} T \\ \vec{a} = Acceleration = (- \hat{i} + 7 \hat{j}) \times 10^{- 6} m / s^{2} \end{array}$
Let the gap be x .