A current I = 18 A is directed along the positive x-axis and perpendicular to a magnetic field. A magnetic force per unit length of 0.27 N/m acts on the conductor in the negative y-direction. Calculate the magnitude and direction of the magnetic field in the region through which the current passes.

Explanation:

Given that,

current I=18A along positive x axis

Magnetic force per length is given as 0.27N/m, acts on negative y axis

F/L=0.27N/m

Magnetic field and direction?

a. The force in magnetic field is given as

F=iLB

Then,

F/L=iB

Since F/L is 0.27N/m and I=18A

So, 0.27=18B

B=0.27/18

B=0.015T

B=15mT

Direction, using the left hand rule

The current is point in positive x and the force motion act in negative direction,

F=L(i×B)

Note rules of cross produce

i×j=k. , j×i=-k

j×k=i. , k×j=-i

k×i=j. , i×k=-j. equation 1

Now applying this to force formula

F=L(i × B), where L is constant

-j=i×B

So comparing this to evaluation 1, this shows that B must k

For this to be negative, B must be k

I.e positive z axis

So the direction of the magnetic field is positive z axis

-4 on

answer( sicar );