A magnetic needle having moment of inertia of 40 g cm 2 has time period of 3 s in earth's horizontal field of 3.6 × 10-5 Wb / m 2. Its magnetic moment will be

Question

A magnetic needle having moment of inertia of 40 g cm 2 has time period of 3 s in earth's horizontal field of 3.6 × 10-5 Wb / m 2. Its magnetic moment will be (A) 0.5 A m 2 (B) 5 A m 2 (C) 0.250 A m 2 (D) 5 × 102 A m 2

Tardigrade Admin · Accepted Answer

T=2 π √(I/M BH) I=40 g cm 2=400 × 10-8 kg m 2 ∴ 3=2 π √(400 × 10-8/36 × 10-6 × M) ⇒ (1/M)=(9/4 π2) × (36/4) ⇒ M=0.5 A m 2

Q. A magnetic needle having moment of inertia of $40 g c m^{2}$ has time period of $3 s$ in earth's horizontal field of $3.6 \times 1 0^{- 5} Wb / m^{2}$ . Its magnetic moment will be

$0.5 A m^{2}$

$5 A m^{2}$

$0.250 A m^{2}$

$5 \times 1 0^{2} A m^{2}$

$T = 2 π \frac{I}{M B _{H}}$
$I = 40 g c m^{2} = 400 \times 1 0^{- 8} k g m^{2}$
$∴ 3 = 2 π \frac{400 \times 1 0 ^{- 8}}{36 \times 1 0 ^{- 6} \times M}$
$\Rightarrow \frac{1}{M} = \frac{9}{4 π ^{2}} \times \frac{36}{4}$
$\Rightarrow M = 0.5 A m^{2}$

Q. A magnetic needle having moment of inertia of 40gcm2 has time period of 3s in earth's horizontal field of 3.6×10−5Wb/m2. Its magnetic moment will be

0.5Am2

5Am2

0.250Am2

5×102Am2