Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
Assumption: Current flows in the same direction.
Using Right hand thumb rule, the direction of the magnetic field at point P due to current I2 is perpendicular to the plane of paper and inwards.
Similarly, at point Q on X2Y2, the direction of magnetic field due to current I1 is perpendicularly outward.
Using Fleming’s left hand rule we can find the direction of forces F12 and F21 which are in opposite directions thus,
By Ampere’s circuited law, we have,
`B^2 =mu_0/(4pi) (2I_2)/d`
Now, F12 = I1LB2 (Where L ≡ length of the conductors)
`F_12 =(mu_0)/(4pi) (2I_1I_2L)/d =mu_0/(2pi)(I_1I_2L)/d`
In similar manner we get,
`F_21 =mu_0/(2pi) (I_1I_2L)/d .... (1)`
From above we get the magnitude of forces F12 and F21 are equal but in opposite direction. So,
F12 = −F21
Therefore, two parallel straight conductors carrying current in the same direction attract each other.
Similarly, we can prove if two parallel straight conductors carry currents in opposite direction, they repel each other with the same magnitude as equation (1).
संबंधित प्रश्न
Find the condition under which the charged particles moving with different speeds in the presence of electric and magnetic field vectors can be used to select charged particles of a particular speed.
Depict the behaviour of magnetic field lines in the presence of a diamagnetic material?
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 \[\vec{E}\] ?
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 \[B = \sqrt{\frac{mg R sin\theta}{v l^2 \cos^2 \theta}}\]
The current generator Ig' shown in figure, sends a constant current i through the circuit. The wire ab has a length l and mass m and can slide on the smooth, horizontal rails connected to Ig. The entire system lies in a vertical magnetic field B. The system is kept vertically in a uniform horizontal magnetic field B that is perpendicular to the plane of the rails (figure). It is found that the wire stays in equilibrium. If the wire ab is replaced by another wire of double its mass, how long will it take in falling through a distance equal to its length?
-
The presence of a large magnetic flux through a coil maintains a current in the coil if the circuit is continuous.
-
A coil of a metal wire kept stationary in a non– uniform magnetic field has an e.m.f induced in it.
-
A charged particle enters a region of uniform magnetic field at an angle of 85° to the magnetic lines of force, the path of the particle is a circle.
-
There is no change in the energy of a charged particle moving in a magnetic field although a magnetic force is acting on it.
A wire carrying current i has the configuration shown in figure. For the magnetic field to be zero at the centre of the circle, θ must be:
A conductor ABOCD moves along its bisector with a velocity 1 m/s through a perpendicular magnetic field of 1 wb/m2, as shown in figure. If all the four sides are 1 m length each, then the induced emf between A and Din approx is ______V.
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).
