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Otrada [13]
4 years ago
9

A 269-turn solenoid is 102 cm long and has a radius of 2.3 cm. It carries a current of 3.9 A. What is the magnetic field inside

the solenoid near its center?
Physics
1 answer:
RUDIKE [14]4 years ago
8 0

Answer:

Magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T

Explanation:

Given;

number of turns of solenoid, N = 269 turn

length of the solenoid, L = 102 cm = 1.02 m

radius of the solenoid, r = 2.3 cm = 0.023 m

current in the solenoid, I = 3.9 A

Magnitude of the magnetic field inside the solenoid near its centre is calculated as;

B = \frac{\mu_o NI}{l} \\\\

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

B = \frac{4\pi*10^{-7} *269*3.9}{1.02} \\\\B = 1.293 *10^{-3} \ T

Therefore, magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T

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A circular coil of radius 5.0 cm and resistance 0.20 Ω is placed in a uniform magnetic field perpendicular to the plane of the c
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Explanation:

Given that,

Assume number of turn is

N= 1

Radius of coil is.

r = 5cm = 0.05m

Then, Area of the surface is given as

A = πr² = π × 0.05²

A = 7.85 × 10^-3 m²

Resistance of

R = 0.20 Ω

The magnetic field is a function of time

B = 0.50exp(-20t) T

Magnitude of induce current at

t = 2s

We need to find the induced emf

This induced voltage, ε can be quantified by:

ε = −NdΦ/dt

Φ = BAcosθ, but θ = 90°, they are perpendicular

So, Φ = BA

ε = −NdΦ/dt = −N d(BA) / dt

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ε = −NA dB/dt

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I = 1.67 × 10^-18 A

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