Magnetism and Matter Test 22

The magnetic susceptibility of ferromagnetic material in paramagnetic region is given by Curie-Weiss Law, which is presented as:

Voltage induced in the coil is given by  $$|\mathcal{E} |= N\dfrac{d\phi}{dt}$$
where $$N$$ is the number of turns in the coil.
So induced voltage will get decreased on decreasing the number of turns in the coil.

Lenz law actually is that induced current always tends to oppose the cause which produce it.So in order to do work against opposing force we have to put extra effort. This extra work leads to periodic change in magnetic flux hence more current is induced. Thus the extra effort is just transformed into electrical energy which is law of conservation of energy.

Lenz's law states - The direction of current induced in a conductor by a changing magnetic field due to Faraday's law of induction will be such that it will create a field that opposes the change that produced it.
The induced EMF produces a current that opposes the change in flux and hence energy.  As the change starts,  induction opposes and slows the change. If the induced EMF were in the same direction as the change in flux, then  that would give us free energy from no apparent source violating conservation of energy. Thus Lenz's law is in accordance with the law of conservation of energy.

The properties of magnetic filed lines are that they start from north pole and end at south pole (outside the magnet). But they flow from south to north pole inside the magnet forming a continuous closed loop. They are very near to each other where the magnetic field is strong whereas they are far away from each other where magnetic field is weak. At poles, magnetic field is the strongest. Thus the magnetic lines of forces are the nearest at the poles. No two magnetic line of forces intersect with each other.

We have, $$W=-MB(\cos {\theta}_2-\cos {\theta}_1)$$
So, $$W_1=-MB(\cos 90^o-\cos 0^o) = MB$$
and $$W_2=-MB(\cos 60^o-\cos 0^o) = \dfrac{1}{2}MB$$ 
As $$W_1 = nW_2$$
$$\therefore n = \dfrac{W_1}{W_2}=\dfrac{MB}{\dfrac{1}{2}MB} = 2$$

Hint:  Here, we have to apply formula of magnetic momentum and angular momentum.

Solution:

Step1: Find magnetic momentum

The magnetic momentum ‘M’ of the particle is given by,

$$M = I \cdot A.....(1)$$

Where, I=Current & A=Area

Also, $$I = \dfrac{q}{t}.....(2)$$

Where, q=charge & t=time

Also, time, $$t = \dfrac{{2\pi }}{\omega }$$

Where, ω=Angular velocity & 2π=Constant

Put above value in equation (2) we get, Current, $$I = \dfrac{{q\omega }}{{2\pi }}$$

We know that,  $$A = \pi {r^2}$$

Now, equation (1) becomes,

$$M = \dfrac{{q\omega }}{{2\pi }} \times \pi {r^2}$$

$$M = \dfrac{{q\omega {r^2}}}{2}.....(3)$$

Step2: Find angular momentum

We know that, Angular momentum, $$L = mvr.....(4)$$

Where, m=Mass, v=velocity & r=Radius

Also, $$v = r\omega $$

Where, r=Radius & ω=Angular velocity

So equation (4) becomes, $$L = m\omega {r^2}.....(5)$$

Step3: Find required ratio

The ratio of the magnitude of its magnetic moment to that of its angular momentum is given by,

$$\dfrac{M}{L} = \dfrac{{\dfrac{{q\omega {r^2}}}{2}}}{{m\omega {r^2}}}$$…..((With reference to equation (3) & (5))

$$ \Rightarrow \dfrac{{q\omega {r^2}}}{2} \times \dfrac{1}{{m\omega {r^2}}}$$

So, $$\dfrac{M}{L} = \dfrac{q}{{2m}}$$

Hence, the required ratio depends on both ‘q’ and ‘m’.

Hence option (C) is correct.

 

 

When the bar is moved toward loop the flux will increase and the induced current will counter this flux.SO when the magnet is above the plane the direction of current will be counter clockwise.