A long, rigid conductor, lying along an xaxis, carries a current of5.0A in the negative direction. A magnetic field Bis present, given by B=(3.0i^+8.0x2j^)mTwith xin meters and Bin milliteslas. Find, in unit-vector notation, the force on the 2.0msegment of the conductor that lies between x=1.0mand x=3.0m.

Short Answer

Expert verified

The force acting on a conductor in unit vector notation is -0.35Nk^

Step by step solution

01

Write the given data

a) The current flowing through the conductor is I=-5.0A

b) The magnetic field is B=3.0i^+8.0x2j^mTor3.0i^+8.0x2j^×10-3T

c) The segment of the conductor on which force is acting as l=2.0m

d) The conductor lies in between is xi=1.0m and xf=3.0m

02

Determine the concept of magnetic force

If a particle is moving with a uniform velocity within a uniform magnetic field, then it experiences a magnetic force due to its charge value that induces a current within the loop. The force acting on the particle is due to the current along the length of the conductive wire. The direction of this magnetic force is given by Fleming's right-hand rule as the force is perpendicular to both the speed and magnetic field acting on it.

The magnetic force along a loop wire is as follows:

dFB=idL×B …… (i)

Here, iis the current in the loop, Bis the magnetic field vector that it experiences, dL is the length vector of the conducting wire.

03

Determine the force acting on a conductor in unit-vector notation

The conductor is lying along the x axis. Hence,dL=dxi^mandB=Bxi^+Byj^T

The expression of the magnetic force on the current carrying wire is given using the above data in equation (i) as follows:

FB=idxi^m×Bxi^+Byj^T

According to the property of the vector product of two vectors,i^×i^=0andi^×j^=k^. The force value can be given using the given data in the above equation as follows:

FB=ixixfBYdxk^

role="math" localid="1662526547252" =-5.0A1.03.08.0×10-3x2Tdxmk^

=-5.0×8.0×x331.03.0k^×10-3N

=-5.0×8.0××10-33.033-1.033k^N

Substitute the values and solve as:

FB=-40××10-327-13k^N=-0.346k^N-0.35k^N

.

Hence, the value of the force is -0.35k^N.

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