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ale4655 [162]
3 years ago
11

There is a uniform magnetic field of magnitude B, pervading all space, perpendicular to the plane of rod and rails. The rod is r

eleased from rest, and it is observed that it accelerates to the left. In what direction does the magnetic field point?
Physics
1 answer:
Charra [1.4K]3 years ago
6 0

The right hand rule to find the direction of the magnetic field for a falling bar is:

  • The charge is positive the magnetic field is outgoing, horizontally and towards us.
  • The charge of the bar is negative, the magnetic field is incoming, that is horizontal away from us.

The magnetic force is given by the vector product of the velocity and the magnetic field.

        F = q v x B

Where the bolds indicate vectors, F is the force, q the charge on the particle, v the velocity and B the magnetic field.

In the vector product, the vectors are perpendicular, which is why the right-hand rule has been established, see attached:

  • The thumb points in the direction of speed.
  • Fingers extended in the direction of the magnetic field.
  • The palm is in the direction of the force if the charge is positive and in the opposite direction if the charge is negative.

They indicate that the bar is dropped, therefore its speed is vertical and downwards, it moves to the left therefore this is the direction of the force, we use the right hand rule, the magnetic field must be horizontal, we have two possibilities:

  • If the charge is positive the magnetic field is outgoing, horizontally and towards us.
  • If the charge of the bar is negative, the magnetic field is incoming, that is, horizontal away from us

In conclusion using the right hand rule we can find the direction of the magnetic field for a falling bar is:

  • The charge of the bar is negative, the magnetic field is incoming, that is horizontal away from us.
  • The charge is positive the magnetic field is outgoing, horizontally and towards us.

Learn more about the right hand rule here:  brainly.com/question/12847190

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You connect a 100-resistor, a 800-mH inductor, and a 10.0-uF capacitor in series across a 60.0-Hz, 120-V (peak) source. The impe
steposvetlana [31]

Answer:

Impedance, Z = 107 ohms

Explanation:

It is given that,

Resistance, R = 100 ohms

Inductance, L=800\ mH=800\times 10^{-3}\ H=0.8\ H

Capacitance, C=10\ \mu F=10\times 10^{-6}\ F=10^{-5}\ F

Frequency, f = 60 Hz

Voltage, V = 120 V

The impedance of the circuit is given by :

Z=\sqrt{R^2+(X_C-X_L)^2}...........(1)

Where

X_C is the capacitive reactance, X_C=\dfrac{1}{2\pi fC}

X_C=\dfrac{1}{2\pi \times 60\times 10^{-5}}=265.65\ \Omega

X_L is the inductive reactance, X_L={2\pi fL}

X_L={2\pi \times 60\times 0.8}=301.59\ \Omega

So, equation (1) becomes :

Z=\sqrt{(100)^2+(265.65-301.59)^2}

Z = 106.26 ohms

or

Z = 107 ohms

So, the impedance of the circuit is 107 ohms. Hence, this is the required solution.

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