CUET · PHYSICS · PYQ PAPER 2025
A short bar magnet of magnetic moment \( 5.25 \times 10^{-2} \text{ J/T} \) is placed with its axis perpendicular to Earth's magnetic field direction. The distance from the center of the magnet on its perpendicular bisector where the resultant field makes 45° with Earth's field will be : (Earth's magnetic field at that place is 0.42 G.)
- A \( 6.3 \times 10^{-2} \text{ m} \)
- B \( 5 \times 10^{-2} \text{ m} \)
- C \( 6.3 \times 10^{-3} \text{ m} \)
- D \( 5 \times 10^{-3} \text{ m} \)
Answer & Solution
Correct Answer
(B) \( 5 \times 10^{-2} \text{ m} \)
Step-by-step Solution
Detailed explanation
\( B_{eq} = B_E \) \( \frac{\mu_0}{4\pi} \frac{M}{r^3} = B_E \) \( r^3 = \frac{\mu_0}{4\pi} \frac{M}{B_E} \) \( r^3 = 10^{-7} \frac{5.25 \times 10^{-2}}{0.42 \times 10^{-4}} \) \( r^3 = 10^{-7} \times 12.5 \times 10^2 \) \( r^3 = 12.5 \times 10^{-5} = 125 \times 10^{-6} \)…
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