Question

Calculate the magnetic field strength needed on a 200-turn square loop 20.0 cm on a side to create a maximum torque of 300 \({\rm{N}} \cdot {\rm{m}}\) if the loop is carrying 25.0 A

Short answer

The magnetic field strength needed to be \( = {\text{1}}{\text{.50}}{\text{T}}\)

Step-by-step solution

Definition of Magnetic Field.

Magnetic forces can be observed in a magnetic field, a vector field in the vicinity of a magnet, an electric current, or a changing electric field.

Strategy

To get the magnetic field, we will use the equation

\(\tau = NIAB\sin (\theta )\) ………………….(1)

and solve it for B

As it was mentioned that the torque is maximum,

\(\begin{aligned}{c}{\rm{sin(\theta )}} = {\rm{sin}}\left( {{\rm{90}}^\circ } \right)\\ = {\rm{1}}\end{aligned}\)

Calculating magnetic field.

Let us calculate the magnetic field

\({\rm{A = 0}}{\rm{.20 m \times 0}}{\rm{.20 m}}\)

\( = {\text{0}}{\text{.04}}{{\text{m}}^{\text{2}}}\)

Therefore, we can evaluate the strength of the magnetic field using equation (1), such that,

\(B = \frac{\tau }{{NIA}}\)

\( = \frac{{{\text{300}}{\text{N}} \cdot {\text{m}}}}{{{\text{200}} \times {\text{25}}{\text{.0}}{\text{A}} \times {\text{0}}{\text{.04}}{{\text{m}}^{\text{2}}}}}\)

\( = {\text{1}}{\text{.50}}{\text{T}}\)

Therefore, the magnetic field strength needed to be \( = {\text{1}}{\text{.50}}{\text{T}}\)