Question

A wire carrying a 30.0-A current passes between the poles of a strong magnet that is perpendicular to its field and experiences a 2.16-N force on the 4.00 cm of wire in the field. What is the average field strength

Answer 1

the average magnetic field strength is 1.8T

Step-by-step explanation:

Using the formula to calculate average magnectic field strength

B = F/(I * L * Sin(θ)

re write the above formula

`B = (F)/(Il\sin \theta)`

Given that F = 2.16N

I = 30 A

L = 4cm = 4 * 10^-2m

θ = 90°

substitute the values

`B = (2.16N)/((30.0A)(0.040m)(\sin 90^0)) \\\\B = 1.80T`

Hence, the average magnetic field strength is 1.8T

Answer 2

The average field strength is 1.8T

Step-by-step explanation:

The magnitude of the wire whose length is L carrying a current I and making an angle θ with a magnetic field of density B is calculated using the following formula.

B = F/(I * L * Sin(θ)

Given that F = 2.16N

I = 30 A

L = 4cm = 4 * 10^-2m

θ = π/2

By substitution

B = 2.16/(30 * 4 * 10^-2 * Sin(π/2))

B = 1.8T

Hence, the average field strength is 1.8T