Name: Electric Field: Time Varying Magnetic Field
Uploaded: 2019-07-19
Duration: 8 min 32 s
Description: Did you know that the cause of the movement of charges in a stationary circular conducting coil placed in a time-varying magnetic field is due to an induced...

Question

A jumper of mass $m$ can slide without friction along parallel horizontal rails separated by a distance $d$ . The rails are connected to a resistor of resistance $R$ and placed in a vertically uniform magnetic field of induction $\vec{B}$ . The jumper is pushed at a velocity $v_{0}$ .

Determine the distance $s$ covered by the jumper before it comes to rest. How does the direction of induction $\vec{B}$ affect the answer?

Accepted Answer

During the motion of the jumper, the magnetic flux through the circuit formed by the jumper, rails, and the resistor changes. An emf is induced in the circuit, and a current is generated. As a result of the action of the magnetic field on the current in the jumper, the latter will be decelerated.

Let us determine the decelerating force $F$ . Let the velocity of the jumper at a certain instant be $v$ . During a short time interval $Δ t$ , the jumper is shifted along the rails by a small distance $v_{0}$ . The change in the area embraced by the circuit is $\vec{B}$ , and the magnetic flux varies by $Δ Φ = B v d Δ t$ during this time. The emf induced in the circuit is

$ε = - \frac{Δ Φ}{Δ t} = - B v d$

According to Ohm's law, the current through the jumper is $I = \frac{E}{R}$ . The force exerted by the magnetic field on the jumper is

$F = Γ B d = - \frac{B^{2} d^{2} v}{R}$

According to Lenz's law, the force $F$ is directed against the velocity $v$ of the jumper. Let us now write the equation of motion for the jumper (over a small distance $Δ x$ ):

$m a = F = - \frac{B^{2} d^{2} v}{R}$

Considering that $a = \frac{Δ v}{Δ t}$ and $v = \frac{Δ x}{Δ t}$ , we obtain

$m Δ v = - \frac{B^{2} d^{2} Δ x}{R}$

It can be seen that the change in the velocity of the jumper is proportional to the change in its $x$ -coordinate (at the initial instant, $x_{0} = 0$ ). Therefore, the total change in velocity $v_{f} = v_{0} = 0 - v_{0} = - v_{0}$ is connected with the change in the coordinate (with the total displacement s) through the relation

$m (- v_{0}) = - \frac{B^{2} d^{2} s}{R}$

Hence we can determine the length of the path covered by the jumper before it comes to rest:

$s = \frac{m R v_{0}}{B^{2} d^{2}}$

When the direction of the magnetic induction $\vec{B}$ forms an angle $α$ with the normal to the plane of the rails, we obtain

$s = \frac{m R v_{0}}{B^{2} d^{2} \cos^{2} α}$

Indeed, the induced emf, and hence the current through the jumper, is determined by the magnetic flux through the circuit, and in this case, the flux is determined by the projection of the magnetic induction $\vec{B}$ on the normal to the plane of the circuit.

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