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

Answer:

the average magnetic field strength is 1.8T

Explanation:

Using the formula to calculate average magnectic field strength

B = F/(I * L * Sin(θ)

re write the above formula

$B = \frac{F}{Il\sin \theta}$

Given that F = 2.16N

I = 30 A

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

θ = 90°

substitute the values

$B = \frac{2.16N}{(30.0A)(0.040m)(\sin 90^0)} \\\\B = 1.80T$

Hence, the average magnetic field strength is 1.8T

Answer 2

Answer:

The average field strength is 1.8T

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