Two bar magnets A and B are identical and arranged as shown. Their lengths are negligible compared to the separation between them. A magnetic needle placed between the magnets at point P gets deflected through an angle \( \theta \) under their influence. The ratio of distances \( d_1 \) and \( d_2 \) is:
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The deflection angle of a magnetic needle between two bar magnets depends on the distances from the magnets and the angle of deflection.
We are tasked with finding the ratio of distances \(d_1\) and \(d_2\) between two identical bar magnets \(A\) and \(B\) arranged as shown, such that a magnetic needle placed at point \(P\) is deflected through an angle \(\theta\). Step 1: Magnetic field due to a bar magnet The magnetic field \(B\) at a distance \(d\) from a bar magnet of magnetic moment \(M\) is given by: \[ B = \frac{\mu_0}{4\pi} \frac{2M}{d^3} \] where \(\mu_0\) is the permeability of free space. Step 2: Magnetic fields at point \(P\) Let the magnetic fields due to magnets \(A\) and \(B\) at point \(P\) be \(B_1\) and \(B_2\), respectively. Since the magnets are identical, their magnetic moments are equal (\(M_A = M_B = M\)). Thus: \[ B_1 = \frac{\mu_0}{4\pi} \frac{2M}{d_1^3} \] \[ B_2 = \frac{\mu_0}{4\pi} \frac{2M}{d_2^3} \] Step 3: Resultant magnetic field at point \(P\) The magnetic needle at point \(P\) experiences a resultant magnetic field \(B_R\) due to the vector sum of \(B_1\) and \(B_2\). The deflection angle \(\theta\) is related to the components of the magnetic fields. The horizontal component of the resultant field is: \[ B_{R_x} = B_1 - B_2 \] The vertical component of the resultant field is: \[ B_{R_y} = 0 \] The tangent of the deflection angle \(\theta\) is given by: \[ \tan \theta = \frac{B_{R_y}}{B_{R_x}} = \frac{0}{B_1 - B_2} = 0 \] This suggests that the deflection angle \(\theta\) is due to the balance of the magnetic fields. Step 4: Equating the magnetic fields For the needle to be deflected by angle \(\theta\), the horizontal components of the magnetic fields must balance: \[ B_1 \cos \theta = B_2 \sin \theta \] Substitute \(B_1\) and \(B_2\): \[ \frac{\mu_0}{4\pi} \frac{2M}{d_1^3} \cos \theta = \frac{\mu_0}{4\pi} \frac{2M}{d_2^3} \sin \theta \] Simplify: \[ \frac{2\cos \theta}{d_1^3} = \frac{2\sin \theta}{d_2^3} \] Rearrange: \[ \frac{d_2^3}{d_1^3} = \frac{2\sin \theta}{2\cos \theta} = 2\tan \theta \] Take the cube root of both sides: \[ \frac{d_2}{d_1} = (2\tan \theta)^{1/3} \] Thus, the ratio of distances is: \[ \frac{d_1}{d_2} = (2\cot \theta)^{1/3} \] Step 5: Match with the options The ratio \(\frac{d_1}{d_2} = (2\cot \theta)^{1/3}\) matches option (1). Final Answer: \(\boxed{1}\)
