A charged particle is projected in a magnetic field B=3i+4j×10-2T. The acceleration of the particle is found to be a=xi+2jm/s2. Find the value of x.

As we have read, Fm⊥B.i.e., the acceleration is a⊥B or a.B=0or xi+2j.3i+4j×10-2=0or 3x+8×10-2=0∴x=-83m/s-2

A charged particle is projected in a magnetic field B=3i+4j×10-2T. The acceleration of the particle is found to be a=xi+2jm/s2. Find the value of x.

As we have read, Fm⊥B.i.e., the acceleration is a⊥B or a.B=0or xi+2j.3i+4j×10-2=0or 3x+8×10-2=0∴x=-83m/s-2

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A charged particle is projected in a magnetic field $B = (3 i + 4 j) \times 10^{- 2} T$ . The acceleration of the particle is found to be $a = (x i + 2 j) m / s^{2}$ . Find the value of $x$ .

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$x = - \frac{4}{3} m / s^{2}$

$x = \frac{6}{3} m / s^{2}$

$x = \frac{8}{3} m / s^{2}$

$x = - \frac{8}{3} m / s^{2}$

answer is D.

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Detailed Solution

As we have read, $F_{m} ⊥ B$ .

i.e., the acceleration is $a ⊥ B$ or $a . B = 0$

or $(x i + 2 j) . (3 i + 4 j) \times 10^{- 2} = 0$

or $(3 x + 8) \times 10^{- 2} = 0$

$∴ x = - \frac{8}{3} m / s^{- 2}$

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A charged particle is projected in a magnetic field B=3i+4j×10-2T. The acceleration of the particle is found to be a=xi+2jm/s2. Find the value of x.

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A charged particle is projected in a magnetic field $B = (3 i + 4 j) \times 10^{- 2} T$ . The acceleration of the particle is found to be $a = (x i + 2 j) m / s^{2}$ . Find the value of $x$ .