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A bar magnet having a magnetic moment of $2 \times 10^{4} J T^{- 1}$ is free to rotate in a horizontal plane. A horizontal magnetic field $B = 6 \times 10^{- 4} T$ exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction $60 °$ from the field is

0.6 J

12 J

6 J

2 J

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Explanation

The work done in rotating a magnetic dipole against the torque acting on it, when placed in magnetic field is stored inside it in the form of potential energy.

When magnetic dipole is rotated from initial position $θ = θ_{1}$ to final position $θ = θ_{2},$ then work done =MB $(\cos θ_{1} - \cos θ_{2})$

=MB $(1 - \frac{1}{2})$

= $\frac{2 \times 10^{4} \times 6 \times 10^{- 4}}{2} = 6 J$

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Curie temperature is the temperature above which -

ferromagnetic material becomes paramagnetic material

paramagnetic material becomes diamagnetic material

paramagnetic material becomes ferromagnetic material

ferromagnetic material becomes diamagnetic material

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Explanation

Ferromagnetism decreases with rise in temperature. If we heat a ferromagnetic substance, then at a definite temperature the ferromagnetic property of the substance suddenly disappears and the substance becomes paramagnetic. The temperature above which a ferromagnetic substance becomes paramagnetic is called the curie temperature of the substance.
Note: The Curie temperature of Iron is $770^{°} C$ and that of Nickel in $358^{°} C$ .

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The current through a choke coil increases from zero to 6 A in 0.3 seconds and an induced emf of 30 V is produced. The inductance of the coil is

5 H

2.5 H

1.5 H

2 H

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Explanation

$\begin{aligned} e = - \frac{LdI}{dt} \\ 30 = L \times \frac{6}{0 .3} \\ L = 1 .5 H \end{aligned}$

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Energy stored in an inductor can be made four times when

inductance is doubled

Current is made $\sqrt{2}$ times

current is doubled

can never to be made

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Explanation

$U = \frac{1}{2} {LI}^{2}$

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A coil wire of a certain radius has 600 turns and a self-inductance of 36 mH. The self-inductance of a second similar coil of 500 turns will be

25 mH

50mH

108mH

75mH

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Explanation

$(1) \frac{L_{1}}{L_{2}} = \frac{N_{1}^{2}}{N_{2}^{2}} \frac{36}{L_{2}} = {(\frac{600}{500})}^{2} L_{2} = 25 mH$

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The current passing through a choke coil of 5 H is decreasing at the rate of 3 As^-1. The emf induced in the coil is

15 V

-15 V

$\frac{5}{3} V$

$\frac{- 5}{3} V$

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Explanation

$e = - L \frac{dl}{dt} = 5 \times 3 = 15 volt (c u r r e n t i s d e c r e a s d i n g)$

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A coil of resistance 20 $Ω$ and inductance 5H has been connected to a 200 V battery. The maximum energy stored in the coil is

250 J

125 J

500 J

100 J

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Explanation

$U = \frac{1}{2} L \frac{E^{2}}{R^{2}} U = \frac{1}{2} \times 5 \times 100 = 250 J$

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A coil has, 1,000 turns and 500 $c m^{2}$ as its area. The plane of the coil is placed at right angles to a magnetic induction field of $2 \times 10^{- 5} W b / m^{2}$ . The coil is rotated through 180 $°$ in 0.2. The average e.m.f. induced in the coil, in milli-volts, is

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Explanation

$e - \frac{d ϕ}{d t} = \frac{2 n B A}{t} e = \frac{2 \times 1000 \times 2 \times 10^{- 5} \times 500 \times 10^{- 4}}{0.2} e = \frac{4 \times 5 \times 10^{- 4}}{2 \times 10^{- 1}} = \frac{20}{2} \times 10^{- 3} = 10 m v$

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A small square loop of wire of side l is placed inside a large square of wire of side L (L>>l). The loops are co-planar and their centres coincide. The mutual inductance of the system is proportional to :

l/L

$l^{2} / L$

L/l

$L^{2} / l$

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Explanation

$ϕ = B A = \frac{4 \times μ i (2 \sin θ)}{4 π L} \times l^{2} M u t u a l i n d u c \tan c e M \propto \frac{l^{2}}{L}$

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Two coils have a mutual inductance of 5 mH. Current changes in the first coil according to the equation I = $I_{0}$ cos wt, where $I_{0}$ = 10 A and $ω = 100 π$ rad/s. Maximum value of e.m.f. induced in the second coil is

5 $π$ volt

2 $π$ volt

4 $π$ volt

$π$ volt

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Explanation

$(1) ε = - M \frac{dl}{dt} ε = - M \frac{d}{dt} (I_{0} \cos ω t) ε = + {MI}_{0} ωsin ωt e = + {MI}_{0} w \sin wt ε_{max} = {Ml}_{0} ω = 5 \times 10^{- 3} \times 10 \times 100 π ε_{max} = {Ml}_{0} ω = 5 \times 10^{- 3} \times 10 \times 100 π = 5 π$

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