What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30º with the direction of a uniform magnetic field of 0.15 T?
Solution and Explanation
Magnitude of the uniform magnetic field, \(B = 0.15 T\)
Angle between the wire and magnetic field, \(θ = 30°\)
Magnetic force per unit length on the wire is given as:\(F = BISinθ\)
\(= 0.15 × 8 ×1 × sin30°\)
\(= 0.6 N m^{–1}\)
Hence, the magnetic force per unit length on the wire is \(0.6 N m^{–1}.\)
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Concepts Used:
Magnetic Force
Magnetic force is the attraction or repulsion force that results from the motion of electrically charged particles. The magnets are attracted or repellent to one another due to this force. A compass, a motor, the magnets that hold the refrigerator door, train tracks, and modern roller coasters are all examples of magnetic power.
A magnetic field is generated by all moving charges, and the charges that pass through its regions feel a force. Depending on whether the force is attractive or repulsive, it may be positive or negative. The magnetism force is determined by the object's charge, velocity, and magnetic field.
Read More: Magnetic Force and Magnetic Field
The magnitude of the magnetic force depends on how much charge is in how much motion in each of the objects and how far apart they are.
Mathematically, we can write magnetic force as:
A charge will feel a force as it passes through a magnetic field at an angle. This force is given by the equation:
A force acts on the motion of charge q traveling with velocity v in a Magnetism field, and this force is:
- Perpendicular to both v and B.
- Perpendicular to sinθ (where θ is the angle between v and B).
- Proportional to the charge q.
- Proportional to the velocity v.