Question

0.23> A bar magnet freely suspended horizontally is found to get aligned in a particular direction only. When slightly disturbed
from its mean position, it performs 30 oscillations per minute. Find the value of horizontal component of earth's magnet field at that
place. The moment of inertia is 1.5 kgm and M = 2 Am.​

Answer 1

Answer:

The horizontal component of earth's magnet field at that

place is 2.6 * 10⁻⁴ T.

Explanation:

Given data:

No. of oscillations per minute = 30

Moment of Inertia, I = 1.5 kgm²

Magnetic moment, M = 2 Am²

We have to find the horizontal component of Earth’s magnetic field, "BΗ".  

We have the formula

Time period, T = 2π√{I / (M * BH) } ……………..(i)

Therefore, we can deduce from (i) i.e.,

No. of oscillations per minute = 1/(2π) * √{(M * BH) / I}

Or, 30 = 1/(2π) * √{(2 * BH) / 1.5}

Or, BH = (30 * 2π * √{1.5 / 2} )²

[ taking value of π = 3.14]

Or, BH = (30 * 2 * 3.14 * 0.866)²

Or, BH = (163.15)²

Or, BH = 26617.9  ≈ 2.6 * 10⁻⁴ T