AP EAMCET · PHYSICS · Magnetic Effects of Current
Two short magnets of equal dipole moments \(M\) are fastened perpendicularly at their centres. The magnitude of the magnetic field at a distance \(d\) from the centre on the bisector of the right angle is ( \(\mu_0=\) Permeability of free space)
- A \(\frac{\mu_0}{4 \pi} \frac{2 \sqrt{2} M}{d^3}\)
- B \(\frac{\mu_0}{4 \pi} \frac{5 M}{d^3}\)
- C \(\frac{\mu_0}{4 \pi} \frac{2 M}{d^3}\)
- D \(\frac{\mu_0}{4 \pi} \frac{10 M}{d^3}\)
Answer & Solution
Correct Answer
(A) \(\frac{\mu_0}{4 \pi} \frac{2 \sqrt{2} M}{d^3}\)
Step-by-step Solution
Detailed explanation
When two short magnets of equal dipole moments \(M\) are fastened perpendicularly at their centres. Then their resultant dipole moment is given as \(\begin{aligned} M^{\prime} & =\sqrt{M_1^2+M_2^2+2 M_1 M_2 \cos \theta} \\ & =\sqrt{M^2+M^2+2 M M \cos 90^{\circ}}\end{aligned}\)…
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