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How strong is the magnetic field inside a solenoid with 10,000turns per meter that carries 20.0 A?

Short Answer

Expert verified

The magnetic field strength inside a solenoid carrying 20.0 A current having 10,000 turns per meter is 0.25 T

Step by step solution

01

Definition of magnetic field

A magnetic field is defined as a position in space near a magnet or an electric current where a physical field is formed by a moving electric charge applying force on another moving electric charge.

02

Given information and formula to used

The number of turns per meter is N=10000

The current in the loop is, I=20.0A

The magnetic field inside a solenoid can be estimated as,

B=0NI鈥︹赌︹赌︹赌︹赌.(1)

Where 0is the magnetic permeability of free space.

03

Step 3: Calculation for Magnetic field inside a solenoid:

Substitute the values in the above expression, we will get

B=410-7N/m21000020.0A=0.25T

Therefore, the magnetic field strength inside a solenoid carrying 20.0A current having 10,000 turns per meter is 0.25T .

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Most popular questions from this chapter

Find the magnitude and direction of the magnetic field at the point equidistant from the wires in Figure 22.58(b), using the rules of vector addition to sum the contributions from each wire.

The force on the rectangular loop of wire in the magnetic field in Figure 22.56 can be used to measure field strength. The field is uniform, and the plane of the loop is perpendicular to the field. (a) What is the direction of the magnetic force on the loop? Justify the claim that the forces on the sides of the loop are equal and opposite, independent of how much of the loop is in the field and do not affect the net force on the loop. (b) If a current of 5.00 A is used, what is the force per tesla on the 20.0-cm-wide loop?


Figure 22.56 A rectangular loop of wire carrying a current is perpendicular to a magnetic field. The field is uniform in the region shown and is zero outside that region.

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