Conducting rod moving on an inclined rail and electromagnetic induction

Explanation

Equations for Induced Electromotive Force and Current

When a conducting rod moves downward along a rail at a speed $v$ , electromagnetic induction occurs when the conducting rod crosses a magnetic wire. The magnitude of the induced electromotive force $V$ in the conducting rod is expressed as follows: $V = vBL$

According to Lenz's law, this electromotive force tries to induce a current in a direction that opposes the change in magnetic flux (counterclockwise when viewed from above). Since the resistance of the resistor is $R$ and other resistances can be ignored, according to Ohm's law, the magnitude of the current $I$ flowing through the circuit is as follows: $I = \frac{V}{R} = \frac{vBL}{R}$

Force Acting on the Conducting Rod and Equation of Motion

A conducting rod through which current $I$ flows is subjected to an Ampere force (electromagnetic force) from the magnetic field. According to Fleming's left-hand rule, the direction of this force is upward along the inclined plane. Its magnitude $F_A$ is expressed as follows: $F_A = IBL = \frac{vB^2L^2}{R}$

The conducting rod is subjected to a gravitational component $mg \sin\theta$ acting parallel to the inclined plane downwards. Therefore, the equation of motion for the conducting rod, with the downward direction of the inclined plane as positive, is as follows: $m \frac{dv}{dt} = mg \sin\theta - \frac{vB^2L^2}{R}$

Calculation of Velocity during Uniform Linear Motion

After a sufficient amount of time has elapsed, the conducting rod will move in a uniform linear motion at a constant velocity. At this time, the acceleration is $0$ ( $\frac{dv}{dt} = 0$ ), so the forces in the direction of the inclined plane are balanced. $mg \sin\theta - \frac{vB^2L^2}{R} = 0$

Solve this equation for $v$ .

$v = \frac{mgR \sin\theta}{B^2L^2}$

Numerical Substitution

We substitute the given constraint values and calculate. Note that $\sin 30.0^\circ = 0.500$ .

$v = \frac{0.200 \times 9.80 \times 4.00 \times 0.500}{0.700^2 \times 1.00^2}$

We calculate the numerator and denominator separately.

Numerator: $0.200 \times 9.80 \times 4.00 \times 0.500 = 3.92$
Denominator: $0.700^2 \times 1.00^2 = 0.490$

Therefore, the velocity $v$ is as follows:

$v = \frac{3.92}{0.490} = 8.00 \text{ m/s}$

The natural number to input is $8$ .

Answer: 8

Conducting rod moving on an inclined rail and electromagnetic induction