A wire having mass per unit length of 0.440 g/cm carries a 2.80 A current horizontally to the south. What are the direction and magnitude of the minimum magnetic field needed to lift this wire vertically upward?
B*I*L = M g = (mass per length)*L*g
L cancels out.
Solve for B and use the right hand rule for the direction of the magnetic force (up)
B = (mass per length)*g/I
= 4.40*10^-2 kg/m*9.8 m/s^2/2.80 A
= ___ Tesla
To find the minimum magnetic field needed to lift the wire vertically upward, we can use the relationship between magnetic field, current, and force in the presence of a magnetic field.
The force experienced by a current-carrying wire in a magnetic field is given by the equation:
F = BIL
Where:
- F is the force acting on the wire,
- B is the magnetic field strength,
- I is the current flowing through the wire, and
- L is the length of the wire.
In this case, the wire is to be lifted vertically upward. This means the force acting on the wire should be directed vertically upward. Therefore, we need to determine the minimum magnetic field strength that can provide an upward force to counteract the gravitational force acting on the wire.
The gravitational force acting on the wire can be calculated using the equation:
F_gravity = mg
Where:
- F_gravity is the gravitational force acting on the wire,
- m is the mass per unit length of the wire, and
- g is the acceleration due to gravity.
Let's calculate the gravitational force acting on the wire:
Given:
- Mass per unit length (m) = 0.440 g/cm
- Current (I) = 2.80 A
First, convert the mass per unit length from grams/centimeter to kilograms/meter:
m = 0.440 g/cm = 0.440 * (1/1000) kg/m = 0.00044 kg/m
Next, we need to determine the force acting on the wire when it is lifted vertically upward. The force can be calculated by multiplying the mass per unit length with the acceleration due to gravity:
F_gravity = (0.00044 kg/m) * 9.8 m/s^2 = 0.004312 N/m
Now, we can rearrange the formula for force (F = BIL) to solve for the magnetic field strength (B):
B = F / (IL)
Substituting the known values, we get:
B = (0.004312 N/m) / (2.80 A * 1 m) = 0.001543 T
Therefore, the minimum magnetic field needed to lift the wire vertically upward is approximately 0.001543 Tesla, directed vertically upward.