JEE Main · 2017 · 02 Apr 2017 · Offline · Elasticity & SHM

A magnetic needle of magnetic moment

6.7 \times 1 0^{- 2}

^{2}

and moment of inertia

7.5 \times 1 0^{- 6}

kg m

^{2}

is performing simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is:

Answer: (C)

Solution

A magnetic needle swinging in a field is just a torsional oscillator, so its period is

T = 2 π I / m B

with

I

the moment of inertia,

m

the magnetic moment and

B

the field. Substituting,

\frac{I}{m B} = \frac{7.5 \times 1 0 ^{- 6}}{( 6.7 \times 1 0 ^{- 2} ) ( 0.01 )} = \frac{7.5 \times 1 0 ^{- 6}}{6.7 \times 1 0 ^{- 4}} = 1.12 \times 1 0^{- 2}

, whose square root is

0.106

. Then

T = 2 π (0.106) = 0.665

s, and ten complete swings take

10 T = 6.65

s. A word of caution: a stray rounding once pushed this to

6.98

s, but carried cleanly the figure settles at

6.65

JEE Main · 2017 · 02 Apr 2017 · Offline · Single correct option