Answer:
9.67 A
Explanation:
We want the magnetic force on the wire to be equal to the weight of a student of mass m=75 kg. The magnetic force on the wire is given by
[tex]F=ILB sin \theta[/tex]
where
I is the current in the wire
L = 2.0 m is the length of the wire
B = 38 T is the magnetic field
[tex]\theta=90^{\circ}[/tex] is the angle between the direction of B and L
While the weight of the student is
[tex]W=mg=(75 kg)(9.8 m/s^2)=735 N[/tex]
where g=9.8 m/s^2 is the acceleration due to gravity.
The problem can be solved by equalizing the two forces: W=F. So we can write
[tex]ILB sin \theta=W[/tex]
And solving for I, the current, we find
[tex]I=\frac{W}{BLsin\theta}=\frac{735 N}{(38 T)(2.0 m)(sin 90^{\circ})}=9.67 A[/tex]
