Question

Three frictionless conducting rails are fixed on an incline plane as shown. The top piece has a resistance R. A free conducting rod of mass m length 1 is placed on the rails, forming a closed loop abcd, and pushed with a force ¥ = -Fî. There is an external magnetic field B = B(cosa î + cosß î + cosy k). (a) Find an expression for the induced e.m.f in the closed loop abcd. Find an expression for the induced current lind and give its direction. Discuss the reason for the direction of this current. (b) Find an expression for the magnetic force Fm which will be acting on the free rod as a result of moving it. (c) Find the components of normal reaction (Nx, Ny, N2), acting on the rod, such that the rod's speed, v, will stay constant. There is no need for an F.B.D. Use vector form of Newton's Law. [15] B = B(cosa î + cosß ſ + cosy k) b y R 444 Free bar || to z-axis Rail ab is || to z-axis F = -Fi Weight of free bar is in - ſ direction e

Answer 1

a) from Faraday's law we know the emf induced in loop is given by

E = do dt (1)

Let's assume the direction of induced current in the loop is anticlockwise if we look from above the plane then unit normal vector of the plane $\hat n$ will be perpendicularly above the inclined plane as shown in figure

न R C. is C व 9 1

U } 기 , N ng on 지 눈

Here we can see.

ñ = sin ei + cos 7

Therefore magnetic flux passes through the loop at its position x' is

$\phi=\vec B.\hat n S$

With S=x' l =area of loop abcd

Put values of B and n and S we get

$\phi=x'lB(cos\alpha\ \sin\theta+\cos\beta\cos\theta)..........(2)$

Put value of Φ in equation (1) we get

$\varepsilon =-lB(cos\alpha\ \sin\theta+\cos\beta\cos\theta)\frac{dx'}{dt}.....................(3)$

Now from figure we can see net force acting on rod along x' direction is

$-m\frac{d^2x'}{dt^2}=(F\cos\theta-mg\sin\theta).....................(4)$here negative sign is taken because with time distance x' reduces. Solve equation (4) we get (we assume that the rod started moving from rest)

$\frac{dx'}{dt}=-\frac{t}{m}(F\cos\theta-mg\sin\theta).....................(5)$

Put this value in equation (3) we get

$\varepsilon =\frac{lBt}{m}(cos\alpha\ \sin\theta+\cos\beta\cos\theta)(F\cos\theta-mg\sin\theta).....................(6)$

This is the expression for induced e.m.f. in the loop.

Since we get the value of induced e.m.f. positive therefore direction of induced current will be along the direction of loop that we have used i.e. anticlockwise.

if we go for physical meaning of the direction of induced current: as we can see net flux passing through the loop abcd has been reducing with time. Therefore direction of magnetic field generated due to induced current must directed along the primary magnetic field and it possible only if induced current is directed in anticlockwise direction.

Therefore in rod direction of induced current is along negative z-axis.

its value will be

$I_m=\frac{\varepsilon}{R} =\frac{lBt}{Rm}(cos\alpha\ \sin\theta+\cos\beta\cos\theta)(F\cos\theta-mg\sin\theta).....................(7)$

Or in vector form

$I_{Ind}=-I_m\hat k$. It is on free rod.

b) therefore magnetic force acting on this free rod is

$\vec F_m=\vec B\times \vec I_{Ind}=-BI_m(\cos\alpha\hat i+\cos\beta\hat j+\cos\gamma\hat k)\times\hat k$

Or

$\vec F_m=-BI_m(\cos\beta\hat i-\cos\alpha\hat j)$.......(8)

C) Rod's speed will remain constant only if net force acting on it is zero.

Therefore net force acting on the rod is

$\vec N+\vec F_m+\vec F+m\vec g=0$

Put all the values we get

$N_x\hat i+N_y\hat j+N_z\hat k- BI_m(\cos\beta\hat i-\cos\alpha\hat j)- F\hat i-mg\hat j=0$

Separate each component and solve it we get

$\\ N_x=BI_m\cos\beta+F\\ \\N_y=mg-BI_m\cos\alpha\\ \\N_z=0$

This is the result for which the rod will move with constant speed.