Answer:
9.67 A
Explanation:
The weight of a student with a mass of m = 75 kg 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.
We want the magnetic force on the wire to be equal to this weight. The magnetic force on the wire is
[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
Since we want W=F, we can write
[tex]ILB sin \theta=W[/tex]
And we can solve it to find the current I:
[tex]I=\frac{W}{BLsin\theta}=\frac{735 N}{(38 T)(2.0 m)(sin 90^{\circ})}=9.67 A[/tex]
