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In hydrogen atom, the electron is making $6.6 \times 10^{15} r e v / s e c$ around the nucleus in an orbit of radius 0.528 Å. The magnetic moment $(A - m^{2})$ will be

$1 \times 10^{- 15}$

$1 \times 10^{- 10}$

$1 \times 10^{- 23}$

$1 \times 10^{- 27}$

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Explanation

$A = {πR}^{2} = 3.142 \times {(0.528)}^{2} \times 10^{- 2} m^{2}$

$\Rightarrow M = i A = 10.5 \times 10^{- 4} \times 3.142 \times {(0.528)}^{2} \times 10^{- 20}$

$= 10 \times 10^{- 24} u n i t s = 1 \times 10^{- 23} u n i t s$

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A particle of charge q and mass m moves in a circular orbit of radius r with angular speed $ω$ . The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on

$ω a n d q$

$ω q a n d m$

$q a n d m$

$ω a n d m$

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Explanation

(c) The effective current $i = \frac{q ω}{2 π} a n d A = {πr}^{2}$ .
Magnetic moment $M = i A = \frac{1}{2} q ω r^{2}$
Angular moment $L = I ω = m r^{2} ω \Rightarrow \frac{M}{L} = \frac{q}{2 m}$

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Two particles each of mass m and charge q are attached to the two ends of a light rigid rod of length 2R. The rod is rotated at constant angular speed about a perpendicular axis passing through its centre. The ratio of the magnitudes of the magnetic moment of the system and its angular momentum about the centre of the rod is:

$\frac{q}{2 m}$

$\frac{q}{m}$

$\frac{2 q}{m}$

$\frac{q}{πm}$

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Explanation

(a) $i = \frac{2 q ω}{2 π} = \frac{q ω}{π}; ∴ M = i A = \frac{q ω}{π} {πR}^{2} = q ω R^{2}$

$L = 2 R . m v = 2 R . m R ω = 2 m R^{2} ω (∴ v = R ω)$

$\Rightarrow \frac{M}{L} = \frac{q}{2 m}$

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If m is the magnetic moment and B is the magnetic field, then the torque is given by

$\vec{m} . \vec{B}$

$\frac{| \vec{m} |}{| \vec{B} |}$

$\vec{m} \times \vec{B}$

$| \vec{m} | . | \vec{B} |$

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Explanation

(c)

$T o r q u e = \vec{m} \times \vec{B}$

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A 250-turn rectangular coil of length 2.1 cm and width 1.25 cm carries a current of 85 $μ$ A and subjected to a magnetic field of strength o.85 T. Work done for rotating the coil by $180^{\circ}$ against the torque is

9.1 $μ J$

4.55 $μ J$

2.3 $μ$ $J$

1.5 $μ J$

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Explanation

(a) Work done for rotating the coil

W=MB( $\cos θ_{1} - \cos θ_{2}$ )

Where, M=maganetic moment

B=maganetic field

Given. $θ_{1} = O^{\circ}, θ_{2} = 180^{\circ}$

$\Rightarrow$ W=MB( $(\cos 0^{°} - \cos 180^{°})$

= 2MB=2 $\times$ NIA $\times B$

= $2 \times 250 \times 85 \times 10^{- 6} {(1.25 \times 2.1 \times 10^{- 4})}_{} \times 85 \times 10^{- 2}$

=9.1 $μ J$ (Approx)

The closest option is (a).

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The resistances of the four arms P, Q,R and S in a Wheatstone's bridge are 10Ω ,30Ω ,30Ω and 90Ω, respectively. The emf and internal resistance of the cell are 7 V and 5 Ω respectively. If the galvanometer resistance is 50Ω, the current drawn from the cell will be

1.0A

0.2A

0.1A

2.0A

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Explanation

(b) Effective resistance,

R_eff=40x120/120+40=4800/160=30Ω

∴ Current I=7/(30+5)=7/35=0.2A

[∴ I=E/R+r]

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A current loop in a magnetic field

experiences a torque whether the field is uniform or non-uniform in all orientations

can be in equilibrium in one orentations.

can be equilibrium in two orientations, both the wquilibriu states are unstable

can be in equilibrium in two orientations, one stable while the other is unstable

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Explanation

(d) For parallel M is stable and for antiparallel is unstable.

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A closely wound solenoid of 2000 turns and area of cross-section $1.5 \times 10^{- 4} m^{2}$ carries a current of $2.0 A .$ It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field $5 \times 10^{- 2} T$ making an angle of $30 °$ with the axis of the solenoid. The torque on the solenoid will be

$3 \times 10^{- 3} N ‐ m$

$1.5 \times 10^{- 3} N ‐ m$

$1.5 \times 10^{- 2} N ‐ m$

$3 \times 10^{- 2} N ‐ m$

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Explanation

$Given, N = 2000, A = 1.5 \times 10^{- 4} m^{2} i = 2.0 A B = 5 \times 10^{- 2} T, and θ = 30 ° Torque, τ = N i B A \sin θ = 2000 \times 2 \times 5 \times 10^{- 2} \times 1.5 \times 10^{- 4} \times \sin 30 ° = 2000 \times 50 \times 10^{- 6} \times \frac{1}{2} = 1.5 \times 10^{- 2} Nm$

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A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in the equilibrium state. The energy required to rotate it by 60^o is W. Now the torque required to keep the magnet in this new position is:

$\frac{W}{\sqrt{3}}$

$\sqrt{3} W$

$\frac{\sqrt{3} W}{2}$

$\frac{2 W}{\sqrt{3}}$

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Explanation

Torque = MBsinθ
= MB sin $60^{0}$ -------(1)
Work done in displacing the magnet from an angle $θ_{1}$ to $θ_{2}$ is -
W = MB(cos $θ_{1}$ – cos $θ_{2}$ )
W = MB(1 – cos $60^{0}$ ) -------(2)
From (1) and (2)
Torque= $\frac{\frac{\sqrt{3} W}{2}}{\frac{1}{} 2} = \sqrt{3} W$

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A 250-Turn rectangular coil of length 2.1 cm and width 1.25 cm carries a current of 85 $μ$ and subjected to the magnetic field of strength 0.85 T. Work done for rotating the coil by 180° against the torque is:

$4.55 μ J$

$2.3 μ J$

$1.15 μ J$

$9.8 μ J$

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Explanation

Given a rectangular coil of length 2.1 cm and width 1.25 cm
Current through coil = 85 $μ A$
No. of turns = 250
B= 0.85 T
Work done, W= MB $(\cos θ_{1} - \cos θ_{2})$
When it is rotated by angle 180⁰ then
W= MB $(\cos 0^{0} - \cos 180^{0}) = M B (1 + 1) = 2 M B$
W = 2(NIA)B
W= $2 \times 250 \times 85 \times 10^{- 6} [1.25 \times 2.1 \times 10^{- 4}] \times 85 \times 10^{- 2}$

W= $9.8 μ J$

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