Physics Electricity

Question

What is the magnetic field in air 0.05 m away from a long straight wire carrying current of 15A?

 

 

 

An air-core solenoid with 2000 loops is 0.60 m long and has a diameter of 0.02 m.  If the current of 5.0 A is sent through it, what will be the flux density within the solenoid?

 

 

 

In the Bohr model of the hydrogen atom, the electron travels with a speed of 2.2 x 10⁶ m/s in a circle of radius of 5.3 x 10-11 m about the nucleus.  Find the value of B at the nucleus due to the electron's motion.

Answer 1

(1)

 

What is the magnetic field in air 0.05 m away from a long straight wire carrying current of 15A?

 

Is this calculus based physics?  If so, have you seen Ampere's law in your class?  I think Ampere's law gives the simplest solution:

 

∫_{closed path} B·dr = μI

 

The integration path is a circle of radius r = 0.05 m centered at the wire and perpendicular to the wire.  Along this path, the magnitude of B is constant, although its direction is changing, since it is tangent to the path.  Then

 

∫_{closed path} B·dr = B(r) (2πr) = μI

 

B(r) = μI / 2πr

 

Look up μ for air in your physics book (very close to μ₀ for a vacuum).  Plug in I = 15A and r = 0.05 m and get B at 0.05 m from the wire.

 

(2)

 

An air-core solenoid with 2000 loops is 0.60 m long and has a diameter of 0.02 m.   If the current of 5.0 A is sent through it, what will be the flux density within the solenoid?

 

The problem is asking for flux density.  The magnetic flux density is the same as B.  Use the formula for a solenoid of N turns and length L:

 

B = μ (N/L) I

 

Find μ for air in tables, and plug in N = 2000, I = 5.0 A, L = 0.60 m.

 

Note:  If the problem asked for magnetic flux (instead of flux density), you would have to multiply B by the cross sectional area of the solenoid:

 

Φ = BA

 

A = π d²/4

 

d = diameter = 0.02 m

 

(3)

 

In the Bohr model of the hydrogen atom, the electron travels with a speed of 2.2 x 10⁶ m/s in a circle of radius of 5.3 x 10^-11 m about the nucleus. Find the value of B at the nucleus due to the electron's motion.

 

Use the Biot-Savart relation for the field B at the center of a circular loop of current.  Treat the electron moving around the nucleus as a circular loop of current.

 

B = (μ₀ / 4π) 2πI / r

 

B = (μ₀ / 2) I / r

 

Find μ₀ in tables; r is given as 5.3 x 10^-11 m.  You still need I.

 

e = charge of electron (disregard its sign).  e = 1.6 x 10^-19 C

 

Get the period T of the electron's orbit from v = 2πr / T.  Both v and r are given, so 

 

T = 2πr / v

 

Then 

 

I = e/T

 

Plug I, μ₀, and r into 

 

B = (μ₀ / 2) I / r

 

to get the magnitude of B.