Question

1. a 0.120 m long copper wire has a mass of 9.02 g and is carrying a current of 5.10 a perpendicular to a uniform magnetic field. the apparatus is placed in a strong magnetic field and the wire is found to levitate. calculate the magnetic field strength. remember to show all work. (4 marks)

Explanation:

Step1: Recall the levitation condition

For the wire to levitate, the magnetic force \( F_B \) must balance the gravitational force \( F_g \). So, \( F_B = F_g \). The formula for magnetic force on a current - carrying wire is \( F_B=ILB\sin\theta \), and since the current is perpendicular to the magnetic field, \( \theta = 90^{\circ} \) and \( \sin\theta=1 \), so \( F_B = ILB \). The gravitational force is \( F_g=mg \), where \( m \) is the mass of the wire and \( g = 9.8\ m/s^2 \) is the acceleration due to gravity.

Step2: Rearrange the formula to solve for \( B \)

From \( ILB=mg \), we can solve for \( B \): \( B=\frac{mg}{IL} \)

Step3: Convert the mass to SI units

The mass \( m = 9.02\ g=9.02\times 10^{- 3}\ kg \), the length \( L = 0.120\ m \), the current \( I = 5.10\ A \), and \( g = 9.8\ m/s^2 \)

Step4: Substitute the values into the formula

\( B=\frac{(9.02\times 10^{-3}\ kg)\times(9.8\ m/s^2)}{(5.10\ A)\times(0.120\ m)} \)
First, calculate the numerator: \( (9.02\times 10^{-3})\times9.8=8.8396\times 10^{-2}\)
Then, calculate the denominator: \( 5.10\times0.120 = 0.612\)
Now, divide the numerator by the denominator: \( B=\frac{8.8396\times 10^{-2}}{0.612}\approx0.144\ T \)

Answer:

The magnetic field strength is approximately \(\boldsymbol{0.144\ T}\) (or depending on the number of significant figures, if we consider the given values: \(m = 9.02\ g\) (3 sig figs), \(L=0.120\ m\) (3 sig figs), \(I = 5.10\ A\) (3 sig figs), the answer can also be presented as \(0.144\ T\) or with more precise calculation as approximately \(0.144\ T\)).