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प्रश्न
Show with the help of a diagram how the force between the two conductors would change when the currents in them flow in the opposite directions?
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उत्तर
Now, let the direction of current in conductor b be reversed. The magnetic field B2 at point P due to current Ia flowing through a will be downwards. Similarly, the magnetic field B1 at point Q due to current Ib passing through b will also be downwards as shown. The force on a will be, therefore, towards the left. Also, the force on b will be towards the right. Hence, the two conductors will repel each other as shown.
संबंधित प्रश्न
If an electric field \[\vec{E}\] is also applied such that the particle continues moving along the original straight line path, what should be the magnitude and direction of the electric field \[\vec{E}\] ?
Two long straight parallel conductors carrying steady currents I1 and I2 are separated by a distance 'd'. Explain briefly, with the help of a suitable diagram, how the magnetic field due to one conductor acts on the other. Hence deduce the expression for the force acting between the two conductors. Mention the nature of this force.
Consider a long, straight wire of cross-sectional area A carrying a current i. Let there be n free electrons per unit volume. An observer places himself on a trolley moving in the direction opposite to the current with a speed \[v = \frac{i}{\text{nAe}}\] and separation from the wire by a distance r. The magnetic field seen by the observer is very nearly
A wire ab of length l, mass m and resistance R slides on a smooth, thick pair of metallic rails joined at the bottom as shown in figure. The plane of the rails makes an angle θ with the horizontal. A vertical magnetic field B exists in the region. If the wire slides on the rails at a constant speed v, show that \[B = \sqrt{\frac{mg R sin\theta}{v l^2 \cos^2 \theta}}\]
Assertion(A): A proton and an electron, with same momenta, enter in a magnetic field in a direction at right angles to the lines of the force. The radius of the paths followed by them will be same.
Reason (R): Electron has less mass than the proton.
Select the most appropriate answer from the options given below:
A deuteron and an alpha particle having equal kinetic energy enter perpendicular into a magnetic field. Let `r_d` and `r_alpha` be their respective radii of the circular path. The value of `(r_d)/(r_alpha)` is equal to ______.
A beam of protons with speed 4 × 105 ms-1 enters a uniform magnetic field of 0.3 T at an angle of 60° to the magnetic field. The pitch of the resulting helical path of protons is close to :
(Mass of the proton = 1.67 × 10-27 kg, charge of the proton = 1.69 × 10-19 C)
A charged particle is accelerated through a potential difference of 12 kV and acquires a speed of 106 ms-1. It is projected perpendicularly into the magnetic field of strength 0.2 T. The radius of the circle described is ______ cm.
A charged particle of charge q and mass m is projected in a region that contains an electric and magnetic field as shown in the figure with velocity V at an angle of 45° with x-direction. If V = `sqrt((qE)/m)`, then net deviation in particle motion will be (neglect the effect of gravity) in a clockwise direction approx ______ °.
An electron (mass 9 × 10−31 kg and charge 1.6 × 10−19 C) moving with speed c/100 (c = speed of light)is injected into a magnetic field `vecB` of magnitude 9 × 10−4 T perpendicular to its direction of motion. We wish to apply an uniform electric field `vecE` together with the magnetic field so that the electron does not deflect from its path. Then (speed of light c = 3 × 108 m s−1).
