Advertisements
Advertisements
Question
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
Solution
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).
RELATED QUESTIONS
A point charge q moving with speed v enters a uniform magnetic field B that is acting into the plane of the paper as shown. What is the path followed by the charge q and in which plane does it move?
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?
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 \[B = \sqrt{\frac{mg R sin\theta}{v l^2 \cos^2 \theta}}\]
Consider the situation shown in figure. The wires P1Q1 and P2Q2 are made to slide on the rails with the same speed 5 cm s−1. Suppose the 19 Ω resistor is disconnected. Find the current through P2Q2 if (a) both the wires move towards right and (b) if P1Q1 moves towards left but P2Q2 moves towards right.
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. Find the velocity of the wire as a function of time.
If an electron is moving with velocity `vecnu` produces a magnetic field `vec"B"`, then ______.
A charged particle moving in a magnetic field experiences a resultant force ______
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 square coil ABCD with its plane vertical is released from rest in a horizontal uniform magnetic field `vec"B"` of length 2L. The acceleration of the coil is ______.
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.
