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NEET PHYSICSMOVING CHARGES AND MAGNETISMMedium

Question

Two similar coils of radius RR are lying concentrically with their planes at right angles to each other. The currents flowing in them are II and 2I2I, respectively. The resultant magnetic field induction at the centre will be:

A

5μ0I2R\frac{\sqrt{5}\mu_0 I}{2R}

B

3μ0I2R\frac{3\mu_0 I}{2R}

C

μ0I2R\frac{\mu_0 I}{2R}

D

μ0IR\frac{\mu_0 I}{R}

Step-by-Step Solution

According to the sources, the magnitude of the magnetic field BB at the centre of a circular loop of radius RR carrying current II is given by the formula B=μ0I2RB = \frac{\mu_0 I}{2R} .

  1. Field from the first coil (B1B_1): For the coil carrying current II, the magnetic field at the centre is B1=μ0I2RB_1 = \frac{\mu_0 I}{2R}.
  2. Field from the second coil (B2B_2): For the coil carrying current 2I2I, the magnetic field at the centre is B2=μ0(2I)2R=2μ0I2RB_2 = \frac{\mu_0 (2I)}{2R} = \frac{2\mu_0 I}{2R}.
  3. Resultant Field (BnetB_{net}): Since the planes of the two coils are at right angles, the magnetic field vectors they produce at the common centre are also perpendicular to each other. The resultant magnetic field is calculated using the Pythagorean theorem for perpendicular vectors: Bnet=B12+B22B_{net} = \sqrt{B_1^2 + B_2^2} .

Substituting the values: Bnet=(μ0I2R)2+(2μ0I2R)2=μ02I24R2(1+4)=5μ0I2RB_{net} = \sqrt{(\frac{\mu_0 I}{2R})^2 + (\frac{2\mu_0 I}{2R})^2} = \sqrt{\frac{\mu_0^2 I^2}{4R^2} (1 + 4)} = \frac{\sqrt{5}\mu_0 I}{2R}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from MOVING CHARGES AND MAGNETISM. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSMOVING CHARGES AND MAGNETISMsimilarradiusconcentricallyplanesangles

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