Bob of a simple pendulum of length $l$ is made of iron . The pendulum is oscillating over a horizontal coil carrying direct current. If the time period of the pendulum is $T$ then

A electron $(q = 1.6 \times 10^{-19}\, C)$ is moving at right angle to the uniform magnetic field $3.534 \times 10^{-5}\, T$. The time taken by the electron to complete a circular orbit is……$µs$

Two very long, straight, parallel wires carry steady currents $I$ and $-I$ respectively. The distance  etween the wires is $d$. At a certain instant of time, a point charge $q$ is at a point equidistant from the two wires, in the plane of the wires. Its instantaneous velocity $v$ is perpendicular to the plane of wires. The magnitude of the force due to the magnetic field acting on the charge at this instant is

In a chamber, a uniform magnetic field of $6.5 \;G \left(1 \;G =10^{-4} \;T \right)$ is maintained. An electron is shot into the field with a speed of $4.8 \times 10^{6} \;m s ^{-1}$ normal to the field.the radius of the circular orbit of the electron is $4.2 \;cm$. obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

$\left(e=1.5 \times 10^{-19} \;C , m_{e}=9.1 \times 10^{-31}\; kg \right)$

The radius of curvature of the path of the charged particle in a uniform magnetic field is directly proportional to