Find Current in a Solenoid Given Magnetic Field and Velocity

In summary: Also, what is the magnitude of the magnetic field in the solenoid given the provided information? Once you have the value of B, you can use the formula I=B/(uN) to calculate the current in the solenoid. Remember to pay attention to units.
  • #1
elitepro
5
0

Homework Statement


An electron circles at a speed of 8510 m/s in a
radius of 2.42 cm in a solenoid. The magnetic
field of the solenoid is perpendicular to the
plane of the electron’s path. Find the current in the solenoid if it has
23.3 turns/cm.
Answer in units of mA


Homework Equations


I think you can use B=uIN
I=B/(uN)
u is the universal vacuum permittivity constant, n is the number of turns

The Attempt at a Solution


I already calculated force to be 2e-6.
There are 56.386 turns total because 23.3*2.42
so I got 2e-6/(4pi*10^-7*56.386)=0.02822
so 28.22 mA
I punched this in and it was wrong.
Did I do something weird?
 
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  • #2
elitepro said:

Homework Statement


An electron circles at a speed of 8510 m/s in a
radius of 2.42 cm in a solenoid. The magnetic
field of the solenoid is perpendicular to the
plane of the electron’s path. Find the current in the solenoid if it has
23.3 turns/cm.
Answer in units of mA

Homework Equations


I think you can use B=uIN
I=B/(uN)
u is the universal vacuum permittivity constant, n is the number of turns

The Attempt at a Solution


I already calculated force to be 2e-6.
There are 56.386 turns total because 23.3*2.42
so I got 2e-6/(4pi*10^-7*56.386)=0.02822
so 28.22 mA
I punched this in and it was wrong.
Did I do something weird?

1. why calculate the force? your first task is to determine B. what relates B to q, m, v and R, all of which are given to you or known?

2. your calculation of no. of turns does not make sense to me. Your formula for B is correct if you replace N, the number of turns, by n, the no. of turns per unit length. n is also given to you.
Be careful to distinguish between N and n. When you have B you can calculate the current from that formula.

Watch your units!
 
  • #3
rude man said:
what relates B to q, m, v and R, all of which are given to you or known?
What equation relates these variables? I can't find one that does
 
  • #4
How about centripetal force in circular motion = magnetic force?
 
  • #5


Your attempt at a solution looks correct. However, it is always a good idea to double check your calculations and make sure that all units are converted correctly. In this case, the force you calculated should be in Newtons (N) and not in milliNewtons (mN). Also, make sure that the units for the magnetic field (B) are in Tesla (T) and not in milliTesla (mT). Once you have the correct units, your final answer should be in Amperes (A) instead of milliAmperes (mA).
 

FAQ: Find Current in a Solenoid Given Magnetic Field and Velocity

1. What is a solenoid?

A solenoid is a long, cylindrical coil of wire that is used to create a magnetic field when an electric current passes through it.

2. How do I calculate the current in a solenoid given the magnetic field and velocity?

The formula for calculating the current in a solenoid is I = BAv, where I is the current, B is the magnetic field, A is the cross-sectional area of the solenoid, and v is the velocity of the charged particles passing through the solenoid.

3. What is the relationship between current and magnetic field in a solenoid?

The current in a solenoid is directly proportional to the magnetic field strength. This means that as the current increases, the magnetic field also increases.

4. What factors can affect the current in a solenoid?

The current in a solenoid can be affected by the number of turns in the coil, the strength of the magnetic field, the cross-sectional area of the solenoid, and the velocity of the charged particles passing through it. Additionally, factors such as temperature and resistance of the wire can also affect the current.

5. How can I use the current in a solenoid for practical applications?

The magnetic field created by the current in a solenoid can be used in various practical applications, such as electromagnets, electric motors, and generators. It can also be used in devices like speakers, doorbells, and MRI machines.

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