Category: Electromagnetic Induction

Q: A thin diamagnetic rod is placed vertically between the poles of an electromagnet . When the current in electromagnet is switched on , then the diamagnetic rod is pushed up , out of horizontal magnetic field . Hence the rod gains gravitational potential energy .The work required to do this comes from

(a) the lattice structure of the material of the rod

(b)the induced electric field due to the changing magnetic field

(c)the magnetic field

(d) the current source

Ans: (d)

Sol: The energy provided by current source is converted into gravitational potential energy .

Q: The magnetic potential energy stored in a certain inductor is 25 mJ , when current in the inductor is 60 mA . This inductor is of inductance

(a) 1.389 H

(b) 13.89 H

(a) 138.88 H

(a) 0.138 H

Ans: (b)

Sol: Magnetic Potential Energy

$ \displaystyle U = \frac{1}{2}L I^2$

$ \displaystyle L = \frac{2 U}{I^2}$

$ \displaystyle L = \frac{2 \times 25 \times 10^{-3}}{60 \times 60 \times 10^{-6}}$

L = 13.89 H

Q: A square loop of side a is placed in the same plane as a long straight wire carrying a current i. The centre of the loop is at a distance r from the wire where r >> a . The loop is moved away from the wire with the a constant velocity v. The induced e.m.f. in the loop is

(a) $ \displaystyle \frac{\mu_0 i a v}{2 \pi r} $

(b) $ \displaystyle \frac{\mu_0 i a^3 v}{2 \pi r^3} $

(a) $ \displaystyle \frac{\mu_0 i v}{2 \pi } $

(d) $ \displaystyle \frac{\mu_0 i a^2 v}{2 \pi r^2} $

Ans: (d)

Sol: Magnetic field intensity at a distance r from the straight wire carrying current is

$ \displaystyle B = \frac{\mu_0}{4\pi} \frac{2i}{r}$

Area of loop, A = a²

Magnetic flux, φ =BA

$ \displaystyle \phi = \frac{\mu_0}{4\pi} \frac{2i}{r} \times a^2$

$\displaystyle \phi = \frac{\mu_0 i a^2}{2\pi r} $

Induced emf in the loop $ \displaystyle e = -\frac{d\phi}{dt} $

$ \displaystyle e = – \frac{d}{dt}(\frac{\mu_0 i a^2}{2\pi r}) $

$ \displaystyle e = – \frac{\mu_0 i a^2}{2\pi} (-\frac{1}{r^2})\frac{dr}{dt} $

$ \displaystyle e = \frac{\mu_0 i a^2}{2\pi r^2}v $

Q: Two coils have self inductance L₁ = 4 mH and L₂ = 1 mH respectively. The currents in the two coils are increased at the same rate. At a certain instant of time, both coils are given the same power. If I₁ ,I₂ are currents in the two coils at that instant of time respectively, then the value of I₁/I2 is

(a) 1/8

(b) 1/2

(c) 1/4

(d) 1/16

Ans: (c)

Sol : Here, L₁ = 4 mH and L₂ = 1 mH

$ \displaystyle \frac{dI_1}{dt} = \frac{dI_2}{dt} $

As P₁ = P₂

e₁ I₁ = e₂ I₂

$ \displaystyle (L_1\frac{dI_1}{dt})I_1 = (L_2 \frac{dI_2}{dt}) I_2 $

L₁ I₁ = L₂ I₂

$ \displaystyle \frac{I_1}{I_2} = \frac{L_2}{L_1} = \frac{1}{4}$

Q: A horizontal straight conductor of mass m and length l is placed in a uniform magnetic field of magnitude B. An amount of charge Q passes through the rod in a very short time such that the conductor begins to move only after all the charge has passage through it. Its initial velocity will be

(a) B Q l m

(b) (B Q l )/m

(c) (B Q )/(l m)

(d) (B l )/(m Q)

Ans: (b)

Sol : If I is the current passing through the rod, then magnetic force on the rod is

F = Bi l

The impulse given by this force to the rod,

$ \displaystyle J = \int F dt $

$ \displaystyle J = \int B l (i dt ) = B l Q $ …(i)

If u is initial velocity acquired by the rod, then

Impulse J = mu …(ii)

From (i) and (ii),

mu = BlQ

u = BlQ/m