Question

In the figure, a 32 cm length of conducting
wire that is free to move is held in place
between two thin conducting wires. All of the
wires are in a magnetic field. When a 6.0 A
current is in the wire, as shown in the figure,
the wire segment moves upward at a constant
velocity.
The acceleration of gravity is 9.81 m/s?.
a) Assuming the wire slides without friction
on the two vertical conductors and has a mass
of 0.13 kg, find the magnitude of the minimum
magnetic field that is required to move the
wire.
Answer in units of T. b) What is the direction?

Answer 1

Answer:

.66354 T

Explanation:

Use F=ILB

B =  $\frac{F}{IL}$

B = Magnetic field

F= force due to magnetic

I= current

L= length in meters

F = mg

Final formula:

B=$\frac{mg}{IL}$

B=$\frac{(.13)(9.8)}{(6)(.32)}$

B= .66354

ayo btw ion know how to find direction, my b G

Answer 2

Answer:

Yes the one above is correct

Explanation:

For the direction, it is out of the page. Yw guys