Question

In a uniform magnetic field of 0.049 T, a magnetic needle performs 20 complete oscillations in 5 seconds as shown. The moment of inertia of the needle is 9.8 × 10⁻⁶ kgm². If the magnitude of magnetic moment of the needle is x × 10⁻⁵ Am²; then the value of ‘x’ is ______.

विकल्प

• 50 π²

• 1280 π²

• 5 π²

• 128 π²

Answer 1

In a uniform magnetic field of 0.049 T, a magnetic needle performs 20 complete oscillations in 5 seconds as shown. The moment of inertia of the needle is 9.8 × 10⁻⁶ kgm². If the magnitude of magnetic moment of the needle is x × 10⁻⁵ Am²; then the value of ‘x’ is 1280 π².

Explanation:

Given: T = `5/20` s⁻¹

I = 9.8 × 10⁻⁶ kgm²

m = x × 10⁻⁵ Am²

B = 0.049 T

Now,

T = `2 pi sqrt(I/(m B)`

⇒ `5/20 = 2 pi sqrt((9.8 xx 10^-6)/(x xx 10^-5 xx 0.049))`

⇒ `5/20 = 2 pi sqrt((200 xx 10^-1)/x)`

Taking squares on both sides,

⇒ `25/400 = 4 pi^2(20/x)`

⇒ `1/16 = 4 pi^2(20/x)`

⇒ x = 16 × 4 π² × 20

⇒ x = 1280 π²