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Two concentric coils each of radius equal to 2pi cm are placed right angles to each other. If 3A and 4A are the currents flowing through the two coils respectively. The magnetic induction (in WB/m^2) at the centre of the coils will be?
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The magnetic induction at the center of the two concentric coils can be calculated using the Biot-Savart law. According to this law, the magnetic field produced by a current-carrying wire at a point is directly proportional to the current and inversely proportional to the distance from the wire.

Let's solve the problem step by step:

1. Calculation of Magnetic Field Due to First Coil:
- The magnetic field at the center of a coil is given by the formula B = μ₀ * I / (2 * R), where B is the magnetic field, μ₀ is the permeability of free space, I is the current, and R is the radius of the coil.
- In this case, the radius of the first coil is 2π cm, so its magnetic field at the center will be B₁ = μ₀ * 3A / (2 * 2π cm).

2. Calculation of Magnetic Field Due to Second Coil:
- Since the two coils are at right angles to each other, the magnetic field produced by the second coil will act perpendicular to the magnetic field of the first coil.
- The magnetic field at the center of the second coil can be calculated using the same formula as the first coil, but with the current value of 4A and the radius of 2π cm.
- Therefore, the magnetic field at the center of the second coil will be B₂ = μ₀ * 4A / (2 * 2π cm).

3. Calculation of Resultant Magnetic Field:
- The magnetic fields produced by the two coils are perpendicular to each other. Since they act at right angles, we can use the Pythagorean theorem to find the resultant magnetic field at the center.
- The resultant magnetic field B can be calculated as B = sqrt(B₁² + B₂²).

4. Conversion into SI Units:
- Finally, to obtain the magnetic induction in WB/m^2, we need to convert the magnetic field from cm to meters and the current from Amperes to Amperes/meter.
- Since 1 cm = 0.01 m and the current is already in Amperes, we can convert B to B_SI = B * 10000 / μ₀.

Let's summarize the calculations:
- B₁ = μ₀ * 3A / (2 * 2π cm)
- B₂ = μ₀ * 4A / (2 * 2π cm)
- B = sqrt(B₁² + B₂²)
- B_SI = B * 10000 / μ₀

By substituting the appropriate values and performing the calculations, you can find the magnetic induction at the center of the coils in WB/m^2.
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Two concentric coils each of radius equal to 2pi cm are placed right angles to each other. If 3A and 4A are the currents flowing through the two coils respectively. The magnetic induction (in WB/m^2) at the centre of the coils will be?
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Two concentric coils each of radius equal to 2pi cm are placed right angles to each other. If 3A and 4A are the currents flowing through the two coils respectively. The magnetic induction (in WB/m^2) at the centre of the coils will be? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two concentric coils each of radius equal to 2pi cm are placed right angles to each other. If 3A and 4A are the currents flowing through the two coils respectively. The magnetic induction (in WB/m^2) at the centre of the coils will be? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two concentric coils each of radius equal to 2pi cm are placed right angles to each other. If 3A and 4A are the currents flowing through the two coils respectively. The magnetic induction (in WB/m^2) at the centre of the coils will be?.
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