Question

1. a motor uses a coil of wire in a magnetic field to generate force. the motor draws a current of 9.50 a through the coil of wire and has a magnetic field of 1.75 t. if the motor is designed to generate 800 n, how long is the wire in the coil assuming that all of the wire creates force? (2 marks)

Explanation:

Step1: Recall the formula for magnetic force on a current - carrying wire

The formula for the magnetic force \( F \) on a current - carrying wire in a magnetic field is \( F = BIL\sin\theta \). When the wire is perpendicular to the magnetic field (which is the case here as we assume all the wire creates force, so we can take \( \sin\theta= 1\)), the formula simplifies to \( F = BIL \). We need to solve for the length \( L \), so we can re - arrange the formula to \( L=\frac{F}{BI} \).

Step2: Identify the given values

We are given that the force \( F = 800\space N \), the magnetic field \( B = 1.75\space T \), and the current \( I=9.50\space A \).

Step3: Substitute the values into the formula

Substitute \( F = 800\space N \), \( B = 1.75\space T \), and \( I = 9.50\space A \) into the formula \( L=\frac{F}{BI} \).

$$ L=\frac{800}{1.75\times9.50} $$

First, calculate the denominator: \( 1.75\times9.50=16.625 \)
Then, calculate the numerator divided by the denominator: \( L=\frac{800}{16.625}\approx48.12\space m \)

Answer:

The length of the wire in the coil is approximately \( \boldsymbol{48.1\space m} \) (or more precisely \( 48.12\space m \))