Three infinetely long wires 1 , 2 and 3 carry equal currents in the directions shown in fiugre. They are placed on the vertices of an equilateral triangle. Let F be the magnitude of force between any two wires. `F_(1)` is the magnitude of force on wire 1 and `F_(2)` the magnitude of force on wire 2. ThenA. `F_(1)=sqrt3F`B. `F_(2)=F`C. Both (a) and (b) are correctD. Both (a) and (b) are wrong

Three infinetely long wires 1 , 2 and 3 carry equal currents in the di

1.	Three infinetely long wires 1 , 2 and 3 carry equal currents in the directions shown in fiugre. They are placed on the vertices of an equilateral triangle. Let F be the magnitude of force between any two wires. `F_(1)` is the magnitude of force on wire 1 and `F_(2)` the magnitude of force on wire 2. ThenA. `F_(1)=sqrt3F`B. `F_(2)=F`C. Both (a) and (b) are correctD. Both (a) and (b) are wrong
Answer» Correct Answer - c

Discussion

No Comment Found

Related InterviewSolutions

A paramagnetic sample shows a net magnetisation of `8Am^-1` when placed in an external magnetic field of `0*6T` at a temperature of `4K`. When the same sample is placed in an external magnetic field of `0*2T` at a temperature of `16K`, the magnetisation will beA. `(32)/(3)Am^(-1)`B. `(2)/(3)Am^(-1)`C. `6Am^(-1)`D. `2.4Am^(-1)`
A charged particle with velocity `hatv=xhati+yhatj` moves in a magnetic field `vecB=yhati+xhatj`. The force acting on the particle has magnitude F. Which one of the following statements is are correct?A. No force will act on particle, if x=yB. `Fprop(x^(2)-y^(2))`if xgtyC. The force will act along z-axis, if xgt yD. The force will act along y-axis, if ygt x
A charged particle moves in a uniform magnetic field `B=(2hati-3hatj)` TA. if velocity of the particles is `(6hatK)m//s`, particle moves in a circleB. if velocity of the particle is `(-4hati+6hatj)m//s`, particle moves in a straight meC. if velocity of the particle is `(hati+2hatj)m//s`, particle moves in a helical pathD. In all the above three cases speed of the particle remains unchanged
A particle of charge q and mass m starts moving from the origin under the action of an electric field `vec E = E_0 hat i and vec B = B_0 hat i` with a velocity `vec v = v_0 hat j`. The speed of the particle will become `2v_0` after a time.A. `t= (2mv_(0))/(qE)`B. `t= (2Bq)/(mv_(0))`C. `t= (sqrt3Bq)/(mv_(0))`D. `t= (sqrt3mv_(0))/(qE)`
A charged particle of unit mass and unit charge moves with velocity `vecv=(8hati+6hatj)ms^-1` in magnetic field of `vecB=2hatkT`. Choose the correct alternative (s).A. the path of the particle may be `x^(2)+y^(2)-4x-21=0`B. the path of the particle may be `x^(2)+y^(2)=25`C. the path of the particle may be `y^(2)+z^(2)=25`D. the time period of the particle will be 3.14 s
An ammeter reads upto 1 ampere. Its internal resistance is 0.81 ohm. To increase the range to 10 A the value of the required shunt isA. `0*03Omega`B. `0*3Omega`C. `0*9Omega`D. `0*09Omega`
Two identical particles having the same mass m and charges +q and -q separated by a distance d enter a uniform magnetic field B directed perpendicular to paper inwards with speeds `v_1 and v_2` as shown in Fig. 1.139. The particles will not collide if A. `dgt(m)/(Bq)(V_(1)-V_(2))`B. `dlt(m)/(Bq)(V_(1)-V_(2))`C. `dgt(2m)/(Bq)(V_(1)-V_(2))`D. `V_(1)=V_(2)`
Let current `i=2A` be flowing in each part of a wire frame as shown in Fig. 1.138. The frame is a combination of two equilateral triangles ACD and CDE of side 1 m. It is placed in uniform magnetic field ` B= 4 T` acting perpendicular to the plane of frame. The magnitude of magnetic force acting on the frame is The pithc of the helical path followed by the particle is p. The radius of the helix will beA. 24 NB. zeroC. 16 ND. 8 N
The CGS unit of magnetising intensity is 1 oersted, where 1 oersted =A. `80Am`B. `80Am^-1`C. `80A^-1m`D. `80A^-1m^-1`
A particle of specific charge (charge/mass) `alpha` starts moving from the origin under the action of an electric field `oversetrarrE = E_(0)hati` and magnetic field `oversetrarrB = B_(0)hatk` Its velocity at `(x_(0),y_(0).0)` is `(4hati +3hatj)` The value of `x_(0)` is .A. `(13)/(2)(alphaE_(0))/(B_(0))`B. `(16alphaB_(0))/(E_(0))`C. `(25)/(2alphaE_(0))`D. `(5alpha)/(2B_(0))`

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment