Magnetic field of the earth is 0.3 G . A magnet is oscillating with the rate of 5 oscillations/min. How much the magnetic field of the earth is increased, so that the number of oscillations becomes 10 per minute?

Question

Magnetic field of the earth is 0.3 G . A magnet is oscillating with the rate of 5 oscillations/min. How much the magnetic field of the earth is increased, so that the number of oscillations becomes 10 per minute? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

f=(1/2 π) √(M BH/I) propto √BH BH propto f2 (BH2/BH1)=((f2/f1))2=((10/5))2=4 BH2=4 BH1 ⇒ BH2=4 × 0.3=1.2 G Δ B=1.2-0.3=0.9 G

Q. Magnetic field of the earth is $0.3 G$ . A magnet is oscillating with the rate of $5$ oscillations/min. How much the magnetic field of the earth is increased, so that the number of oscillations becomes $10$ per minute?

$f = \frac{1}{2 π} \frac{M B _{H}}{I} \propto B_{H}$
$B_{H} \propto f^{2}$
$\frac{B _{H_{2}}}{B _{H_{1}}} = (\frac{f _{2}}{f _{1}})^{2} = (\frac{10}{5})^{2} = 4$
$B_{H_{2}} = 4 B_{H_{1}}$
$\Rightarrow B_{H_{2}} = 4 \times 0.3 = 1.2 G$
$Δ B = 1.2 - 0.3 = 0.9 G$

Q. Magnetic field of the earth is 0.3G . A magnet is oscillating with the rate of 5 oscillations/min. How much the magnetic field of the earth is increased, so that the number of oscillations becomes 10 per minute?

f=2π1​IMBH​​​∝BH​​ BH​∝f2 BH1​​BH2​​​=(f1​f2​​)2=(510​)2=4 BH2​​=4BH1​​ ⇒BH2​​=4×0.3=1.2G ΔB=1.2−0.3=0.9G

Q. Magnetic field of the earth is $0.3 G$ . A magnet is oscillating with the rate of $5$ oscillations/min. How much the magnetic field of the earth is increased, so that the number of oscillations becomes $10$ per minute?

$f = \frac{1}{2 π} \frac{M B _{H}}{I} \propto B_{H}$
$B_{H} \propto f^{2}$
$\frac{B _{H_{2}}}{B _{H_{1}}} = (\frac{f _{2}}{f _{1}})^{2} = (\frac{10}{5})^{2} = 4$
$B_{H_{2}} = 4 B_{H_{1}}$
$\Rightarrow B_{H_{2}} = 4 \times 0.3 = 1.2 G$
$Δ B = 1.2 - 0.3 = 0.9 G$