Electromagnetic Force | MCQs

Explanation

When a wire is placed perpendicular to a magnetic field, the force on the wire is proportional to the current flowing through it.

If the current increases, the force on the wire will also increase.

Explanation

The force $F$ acting on a charged particle moving in a magnetic field is given by the Lorentz force equation, which can be expressed as:

$$F = q(mathbf{v} times mathbf{B})$$

Where:

• $\mathrm{<font\; color="\#0000ff"><b>F\; is\; the\; force\; acting\; on\; the\; particle,</b></font>}$
• $q$ is the charge of the particle,
• $mathbf{v}$ is the velocity vector of the particle,
• $\mathbf{B}$ is the magnetic flux density (magnetic field strength) vector.

The cross product $\mathbf{v}\times \mathbf{B}$ indicates that the direction of the force is perpendicular to both the velocity and the magnetic field vectors. Thus, the correct expression for the force is $F=qv\times B$.

Explanation

When an electron moves perpendicular to a magnetic field, the electromagnetic force (also known as the Lorentz force) acting on the electron is given by the formula:

F=qvBsin(θ)

Where:

• $F$ is the force,
• $q$ is the charge of the electron,
• $v$ is the velocity of the electron,
• $B$ is the magnetic field strength,
• $$θtheta is the angle between the velocity vector and the magnetic field.

When the electron is moving perpendicular to the magnetic field, $theta = 90^circ$ and $sin(90^circ) = 1$

Therefore, the force becomes:

$$F=qvB$$

This means the force is at its maximum value since both the charge and the velocity contribute to the force. Thus, the statement that the force acting is maximum is correct.