A wire of 62cm length and mass 13g is suspended by a pair of flexible leads in a magnetic field of 0.44T. What are the magnitude and direction of the current required to remove the tension in the supporting leads?
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force=.013*9.8=(I)(.62)(.44)
solve for I in amperes
I=.013*9.8/(.62*.44) amps
To determine the magnitude and direction of the current required to remove the tension in the supporting leads, we need to consider the interaction between the magnetic field and the current-carrying wire.
First, let's calculate the weight of the wire.
Weight = mass * acceleration due to gravity
Weight = 0.013 kg * 9.8 m/s^2 (taking g as 9.8 m/s^2)
Weight = 0.1274 N
Next, we consider the tension in the supporting leads. When the wire is at rest, the weight is balanced by the tension in the leads. This tension force is equal to the weight of the wire.
Now, let's consider the magnetic force acting on the wire. The magnetic force on a current-carrying wire is given by the formula:
Magnetic Force = current * length * magnetic field strength * sin(theta)
Where:
- current is the magnitude of the current flowing through the wire (in amperes)
- length is the length of the wire (in meters)
- magnetic field strength is the strength of the magnetic field (in teslas)
- theta is the angle between the current direction and the magnetic field direction
In this case, the length of the wire is 0.62 m, and the magnetic field strength is 0.44 T.
To remove the tension in the supporting leads, we need to make the magnetic force on the wire equal to zero. Since the direction of the magnetic force is perpendicular to the direction of the current and the magnetic field, we need to find the angle at which the magnetic force is perpendicular to the tension force.
Since the weight and magnetic force are acting in opposite directions, the tension force can be represented as a right triangle. The angle between the magnetic force and the tension force is the same as the angle between the magnetic force and the vertical direction.
Let's find this angle:
sin(theta) = Weight / (current * length * magnetic field strength)
sin(theta) = 0.1274 / (current * 0.62 * 0.44)
To make the magnetic force equal to zero, sin(theta) should be equal to 1. Therefore:
1 = 0.1274 / (current * 0.62 * 0.44)
current = 0.1274 / (1 * 0.62 * 0.44)
current = 0.1274 / 0.27016
current ≈ 0.471 A
So, the magnitude of the current required to remove the tension in the supporting leads is approximately 0.471 A.
For the direction of the current, it follows the right-hand rule, which states that if you point your right thumb in the direction of the current, your fingers will wrap around the wire in the direction of the magnetic field. Since the magnetic field is not specified in this question, we cannot determine the exact direction of the current without additional information.
Please let me know if there is anything else I can help you with!