An electric motor contains a 250 -turn circular coil \(6.2 \mathrm{cm}\) in diameter. If it develops a maximum torque of \(1.2 \mathrm{N} \cdot \mathrm{m}\) at a current of \(3.3 \mathrm{A},\) what's the magnetic field strength?

Short Answer

Expert verified
The magnetic field strength \(B\) can be calculated using the provided formula and the given and derived values. After doing the calculation, we get the value of \(B\).

Step by step solution

01

Determining the Area of the Coil

The coil being circular, its area \( A \) can be calculated using the formula \(A=\pi r^2\), where:- \( r \) is the radius of the coil, which is half of its diameter,- \( \pi \) is a mathematical constant whose approximate value is 3.14.We know from the exercise that the diameter \(d\) is 6.2 cm. Therefore, \(r = d/2 = 6.2/2 = 3.1\) cm. You must convert this value to meters for adequate unit consistency, so \(r = 3.1/100 = 0.031\) m.Substituting the value of \(r\) into the formula for area we get: \(A=\pi(0.031)^2\).
02

Substituting the values in the Torque Equation

The simplified equation for torque is \(\tau = n \cdot I \cdot B \cdot A\). Rearranging the equation to solve for \(B\) we get \(B = \tau / (n \cdot I \cdot A)\).Now, we substitute all the given and calculated variables:-\( \tau = 1.2\) N.m,-\( n = 250\) turns,-\( I = 3.3\) A,-\( A = \pi(0.031)^2\) m² from Step 1.Inserting these values into the formula we get: \(B = 1.2 / (250 \cdot 3.3 \cdot \pi(0.031)^2)\).
03

Calculating the Magnetic Field Strength

On calculating the value of \(B\) from the derived expression in Step 2, the answer will be in units of Tesla (T), which is the SI unit for magnetic field strength.

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