Question

a magnetic needle of pole strength 20√3 A-m is pivoted at its centre. It's north pole is pulled eastward by a string. The horizontal force required to produce a deflection of 30° from magnetic meridian (take Bh = 10^-4 T) is​

Answer 1

Explanation:

Given A magnetic needle of pole strength 20√3 A-m is pivoted at its centre. It's north pole is pulled eastward by a string. The horizontal force required to produce a deflection of 30° from magnetic meridian

• Given a magnetic needle of pole strength m = 20 √3 A-m.
• Also deflection from magnetic meridian theta = 30 degree
• Given B h = 10^-4 T
• Now the magnetic meridian B h = B cos theta
•                                   Or B = B h / cos theta
•                                          = 10^-4 / cos 30
•                                          = 10^-4 / √3 / 2
•                                           = 2 x 10 ^-4 / 1.732  
•                                           = 1.1547 x 10^-4 T
•                        Now F = m B
•                              So F = 20 √3 x 1.1547 x 10^-4 N
•                                    F = 39.9 x 10^-4 N
•                        Or F = 4 x 10^-3 N

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https://brainly.in/question/14585399