In a uniform magnetic field, the magnetic needle has a magnetic moment $9.85\times 10^{-2}\mathrm{~A}/\mathrm{m}^{2}$ and moment of inertia $5\times 10^{-6}\mathrm{~kg~m}^{2}$. If it performs 10 complete oscillations in 5 seconds then the magnitude of the magnetic field is ______ mT. [Take $\pi^{2}$ as 9.85]

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Magnetic Properties of Matter

In a uniform magnetic field, the magnetic needle has a magnetic moment

9.85 \times 1 0^{- 2} A / m^{2}

and moment of inertia

5 \times 1 0^{- 6} kg m^{2}

. If it performs 10 complete oscillations in 5 seconds then the magnitude of the magnetic field is ______ mT. [Take

π^{2}

as 9.85]

Answer: 8

Ten complete oscillations in

5

s give a period

T = 0.5

s. A magnetic needle of moment

M

and moment of inertia

I

oscillating in a field

B

has

T = 2 π \frac{I}{M B} \Rightarrow B = \frac{4 π ^{2} I}{M T ^{2}}

Substituting

I = 5 \times 1 0^{- 6} kg m^{2}

M = 9.85 \times 1 0^{- 2}

T = 0.5

s and

π^{2} = 9.85

B = \frac{4 \times 9.85 \times 5 \times 1 0 ^{- 6}}{9.85 \times 1 0 ^{- 2} \times 0.25} = \frac{2 \times 1 0 ^{- 5}}{2.5 \times 1 0 ^{- 3}} = 8 \times 1 0^{- 3} T

The magnitude of the field is

8

mT.

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Integer answer