The magnetic field due to circular coil of 200 turns of diameter 0.1m carrying a current of 5A at a point on the axis of the coil at a distance 0.15m from the center of the coil will bea)b)c)d)Correct answer is option 'D'. Can you explain this answer?

Question

Answer

B=μ₀2πnIa²/4π (a²+x²)^3/2
=10⁻⁷×2×(22/7)×200×5×(0⋅1/2)²/ [(0⋅1/2)²+(0⋅15)²]^3/2
=39.74x10^-5

Answer

Xhxhudysa sjxkxhxh htw54276/,*@*$*@6+4*5/6/5-9'8/53$9=8*5+6/834+5@6''#$)'45*6:/:=

Answer

B=μ₀2πnIa²/4π (a²+x²)^3/2
=10⁻⁷×2×(22/7)×200×5×(0⋅1/2)²/ [(0⋅1/2)²+(0⋅15)²]^3/2
=39.74x10^-5

Answer

B=(u.NIR^2)/2(R^2+x^2)^3/2use this formula...you will get the answer

Answer

Given:
Number of turns (N) = 200
Diameter of the coil (d) = 0.1 m
Radius of the coil (r) = d/2 = 0.1/2 = 0.05 m
Current flowing through the coil (I) = 5 A
Distance from the center of the coil to the point on the axis (x) = 0.15 m

To find:
The magnetic field at the given point on the axis of the coil.

Formula:
The magnetic field due to a circular coil at a point on its axis is given by the formula:
B = (μ₀ × N × I × R²) / (2 × (R² + x²)^(3/2))

Where:
B is the magnetic field
μ₀ is the permeability of free space (4π × 10^-7 T·m/A)
N is the number of turns
I is the current flowing through the coil
R is the radius of the coil
x is the distance from the center of the coil to the point on the axis

Calculation:
Substituting the given values into the formula, we have:
B = (4π × 10^-7 T·m/A) × 200 × 5 × (0.05)² / (2 × (0.05)² + (0.15)²)^(3/2)

Simplifying the equation, we get:
B = (4π × 10^-7 T·m/A) × 200 × 5 × (0.0025) / (2 × 0.0025 + 0.0225)^(3/2)
B = (4π × 10^-7 T·m/A) × 200 × 5 × 0.0025 / (0.005 + 0.0225)^(3/2)
B = (4π × 10^-7 T·m/A) × 200 × 5 × 0.0025 / 0.0275^(3/2)
B = (4π × 10^-7 T·m/A) × 200 × 5 × 0.0025 / 0.0025^(3/2)
B = (4π × 10^-7 T·m/A) × 200 × 5
B = (4π × 10^-7 T·m/A) × 1000

Now, substitute the value of π (pi) as approximately 3.14:
B = (4 × 3.14 × 10^-7 T·m/A) × 1000
B = 1.256 × 10^-6 T·m/A × 1000
B = 1.256 × 10^-6 T·m/A × 10^3
B = 1.256 × 10^-3 T·m/A

Therefore, the magnetic field due to the circular coil at the given point on the axis is 1.256 × 10^-3 T·m/A, which is equivalent to 39.74 × 10^-5 T (option D).

Answer

B=μ₀2πnIa²/4π (a²+x²)^3/2
=10⁻⁷×2×(22/7)×200×5×(0⋅1/2)²/ [(0⋅1/2)²+(0⋅15)²]^3/2
=39.74x10^-5

Top Courses for Class 12View all

English Class 12

Chemistry Class 12

Physics Class 12

English Flamingo Class 12

Physical Education Class 12(XII) - Notes & Model Test Papers

Top Courses for Class 12

Top Courses for Class 12

Related Content

Schools

Competitive Examinations

Quick Access Links

Technical Exams