English
Karnataka Board PUCPUC Science Class 11

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

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Question

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
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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 34: Magnetic Field - Exercises [Page 230]

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HC Verma Concepts of Physics Volume 1 and 2 [English]
Chapter 34 Magnetic Field
Exercises | Q 4 | Page 230
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