Question

A short magnet oscillates with a time period of 0.1 s at a place where the horizontal magnetic field is 24 \(\mu T\). A downward current of 18 A is established in a vertical wire kept at a distance of 20 cm east of the magnet. The new time period of oscillations of the magnet is:

• A. \( 0.1 \, s \)
• B. \( 0.089 \, s \)
• C. \( 0.076 \, s \)
• D. \( 0.057 \, s \)

Hint:

The time period of a magnetic dipole oscillating in a magnetic field is inversely proportional to the square root of the magnetic field. If an external current contributes to the field, use vector addition to find the effective field.

Answer 1

The Correct Option is C

Step 1: Understanding the Magnetic Field Contribution 
The original oscillation time period of the magnet is given by: \[ T_0 = 0.1 \, s \] The total effective horizontal magnetic field \( B_{\text{eff}} \) at the magnet due to the combination of the Earth's horizontal magnetic field \( B_H \) and the magnetic field \( B_I \) due to the vertical current is given by: \[ B_{\text{eff}} = \sqrt{B_H^2 + B_I^2} \] where: - \( B_H = 24 \, \mu T \) - \( B_I \) is the field due to the wire, given by: \[ B_I = \frac{\mu_0 I}{2 \pi d} \] where: - \( \mu_0 = 4\pi \times 10^{-7} \), - \( I = 18 \) A, - \( d = 20 \) cm \( = 0.2 \) m. 

Step 2: Calculating \( B_I \) \[ B_I = \frac{4\pi \times 10^{-7} \times 18}{2 \pi \times 0.2} \] \[ B_I = \frac{72 \times 10^{-7}}{4 \times 10^{-1}} \] \[ B_I = 18 \times 10^{-6} T = 18 \, \mu T \] 

Step 3: Finding \( B_{\text{eff}} \) \[ B_{\text{eff}} = \sqrt{(24)^2 + (18)^2} \] \[ B_{\text{eff}} = \sqrt{576 + 324} = \sqrt{900} = 30 \, \mu T \] 

Step 4: Finding the New Time Period 
Since the time period of oscillation is inversely proportional to the square root of the magnetic field: \[ T' = T_0 \sqrt{\frac{B_H}{B_{\text{eff}}}} \] \[ T' = 0.1 \times \sqrt{\frac{24}{30}} \] \[ T' = 0.1 \times \sqrt{0.8} \] \[ T' = 0.1 \times 0.894 \] \[ T' \approx 0.076 \, s \]