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An Experimenter'S Diary Reads as Follows: "A Charged Particle is Projected in a Magnetic Field of ( 7.0 → I − 3.0 → J ) × 10 − 3 T. the Acceleration of the - Physics

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An experimenter's diary reads as follows: "A charged particle is projected in a magnetic field of `(7.0 vec i - 3.0 vecj)xx 10^-3 `T. The acceleration of the particle is found to be `(x veci + 7.0 vecj )` The number to the left of i in the last expression was not readable. What can this number be?

Sum

Solution

Given,

Magnetic field, B = (7.0i − 3.0j) × 10−3 T

Acceleration of the particle, a = (xi + 7j) × 10−6 m/s2

We have denoted the unidentified number as x.

B and a are perpendicular to each other. (Because magnetic force always acts perpendicular to the motion of the particle)

So, the dot product of the two quantities should be zero.

That is, B.a = 0

⇒ 7x × 10−3 × 10−6 − 3 × 10−3 × 7 × 10−6 = 0

⇒ 7x − 21 = 0

`x = 21/7 = 3`

Acceleration of the particle is (3i + 7j) × 10−6 m/s2.

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Force on a Moving Charge in Uniform Magnetic and Electric Fields
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Chapter 12: Magnetic Field - Exercises [Page 230]

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HC Verma Concepts of Physics Vol. 2 [English] Class 11 and 12
Chapter 12 Magnetic Field
Exercises | Q 4 | Page 230

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