Question

A wire carrying a 27.5 A current passes between the poles of a strong magnet perpendicular to its field and experiences a 2.19 N force on the 4.00 cm of wire in the field. What is the average field strength (in T)

Answer 1

B=1.99 T

Step-by-step explanation:

Given that

Current ,I = 27.5 A

Force ,F= 2.19 N

Length ,L = 4 cm = 0.0 4 m

We know that force on a magnet is given as

F= I B L sin θ

B=Magnetic filed

F=Force

L=Length

I=Current

Given that current and force is perpendicular to each other that is why

θ = 90 °

2.19 = 27.5 x 0.04 x B x sin 90 ° ( sin 90 ° = 1 )

`B=(2.19)/(27.5* 0.04)\ T`

B=1.99 T

Therefore the value of magnetic field will be 1.99 T.