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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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Two long parallel copper wires carry currents of 5A each in opposite directions. If the wires are separated by a distance of 0.5 m, then the force between the two wires is :

$10^{- 5} N,$ attractive

$10^{- 5} N,$ repulsive

$2 \times 10^{- 5} N,$ attractive

$2 \times 10^{- 5} N,$ repulsive

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Explanation

(b) $F = 10^{- 7} \frac{2 i_{1} i_{2}}{a} = 10^{- 7} \times \frac{2 \times 2 \times 5}{0.5} = 10^{- 5} N$ (repulsive)

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A long wire A carries a current of 10 amp. Another long wire B, Which is parallel to A and separated by 0.1m from A, carries a current of 5 amp, in the opposite direction to that in A. what is the magnitude and nature of the force experienced per unit length of B $(μ_{0} = 4 π \times 10^{- 7} w e b e r / a m p - m)$

Repulsive force of $10^{- 4} N / m$

Attractive force of $10^{- 4} N / m$

Repulsive force of $2 π \times 10^{- 5} N / m$

Attractive force of $2 π \times 10^{- 5} N / m$

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Explanation

(a) $F = \frac{μ_{0}}{4 π} \frac{2 i_{1} i_{2}}{a} = 10^{- 7} \times \frac{2 \times 10 \times 5}{0.1} = 10^{- 4} N$ (Repulsive)

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The relation between voltage sensitivity ( $σ_{v}$ ) and current sensitivity ( $σ_{i}$ ) of a moving coil galvanometer is (Resistance of Galvanometer = G)

$\frac{σ_{i}}{G} = σ_{v}$

$\frac{σ_{v}}{G} = σ_{i}$

$\frac{G}{σ_{v}} = σ_{i}$

$\frac{G}{σ_{i}} = σ_{v}$

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Explanation

(a) $σ_{i} = \frac{θ}{i} = \frac{θ}{i G} \cdot G = σ_{ν} G \Rightarrow \frac{σ_{i}}{G} = σ_{v}$

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Two galvanometers A and B require 3mA and 5mA respectively to produce the same deflection of 10 divisions. Then

A is more sensitive than B

B is more sensitive than A

A and B are equally sensitive

Sensitiveness of B is 5/3 times that of A

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Explanation

(a) Sensitivity (S) $= \frac{θ}{i} \Rightarrow \frac{S_{A}}{S_{B}} = \frac{i_{B}}{i_{A}} = \frac{5}{3} \Rightarrow S_{A} > S_{B}$

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Two long conductors, separated by a distance d carry current $I_{1}$ and $I_{2}$ in the same direction. They exert a force F on each other. Now the current in one of them is increased to two times and its direction is reversed. The distance is also increased to $3 d$ . The new value of the force between them is-

$- 2 F$

$2 F / 3$

$- 2 F / 3$

$- F / 3$

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Explanation

$F \propto \frac{i_{1} i_{2}}{a}$ ; Since one of the current increase two times and distance increases three times, so force

become $\frac{2}{3}$ times. Also due to the reversal of direction of current, force becomes negative.

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Two thin, long, parallel wires, separated by a distance ‘d’ carry a current of ‘i’ A in the same direction. They will

Attract each other with a force of $μ_{0} i^{2} / (2 {πd}^{2})$

Repel each other with a force of $μ_{0} i^{2} / (2 {πd}^{2})$

Attract each other with a force of $μ_{0} i^{2} / (2 πd)$

Repel each other with a force of $μ_{0} i^{2} / (2 πd)$

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Explanation

(c) $(\frac{F}{l}) = \frac{μ_{0}}{4 π} . \frac{2 i_{1} i_{2}}{a} \Rightarrow (\frac{F}{l}) = \frac{μ_{0}}{4 π} . \frac{2 i^{2}}{d} = \frac{μ_{0} i^{2}}{2 πd}$ (Attractive)

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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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A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the center of loop is B. It is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be

$n^{2} B$

2nB

$2 n^{2} B$

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Explanation

(b)

$B_{c e n t r e} = n \times (\frac{μ_{0} i}{2 R}) w h e r e n i s t h e n u m b e r o f t u r n s C a s e 1 : B = \frac{μ_{0} i}{2 R} C a s e 2 : L e t r a d i u s b e r a n d n u m b e r o f t u r n s b e n 2 πr \times n = 2 πR New radius r = \frac{R}{n} Magnetic field B_{1} = n \times (\frac{μ_{0} i}{2 r}) = n \times (\frac{μ_{0} i}{2 \frac{R}{n}}) = \frac{n^{2} \times μ_{0} i}{2 R} B_{1} = n^{2} \times B$

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