Magnetic Dipole and Dipole Moment and Torque on a Current-Carrying Planar Loop in Uniform Magnetic Field

Q 11 :

A square loop of area 25 ${cm}^{2}$ has a resistance of $10 Ω$ . The loop is placed in uniform magnetic field of magnitude 40.0 T. The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in 1.0 sec, will be [2023]

$2.5 \times 10^{- 3}$ J
$1.0 \times 10^{- 3}$ J
$1.0 \times 10^{- 4}$ J
$5 \times 10^{- 3}$ J

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4

(2)

$l = 50 cm$

$t = 1 s$

$∴$ $V = \frac{0.05}{1} = 0.05 m/s$

$i = \frac{40 \times 0.05 \times 0.05}{10} = 0.01 A$

$∴$ $F = B_{I ∆ C} ℓ = 40 \times 0.01 \times 0.05$

$F = 0.02 N$

$∴$ $W = F \times ℓ = 0.02 \times 0.05$

$∴$ $W = 1 \times 10^{- 3} J$

Q 12 :

The magnetic moments associated with two closely wound circular coils A and B of radius $r_{A} = 10$ cm and $r_{B} = 20$ cm respectively are equal if (where $N_{A}, I_{A}$ and $N_{B}, I_{B}$ are number of turns and current of A and B respectively). [2023]

$2 N_{A} I_{A} = N_{B} I_{B}$
$N_{A} = 2 N_{B}$
$N_{A} I_{A} = 4 N_{B} I_{B}$
$4 N_{A} I_{A} = N_{B} I_{B}$

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4

(3)

$M = N I A$

$M_{A} = M_{B}$

$∴$ $N_{A} I_{A} A_{A} = N_{B} I_{B} A_{B}$

$∴$ $N_{A} I_{A} π {(0.1)}^{2} = N_{B} I_{B} π {(0.2)}^{2}$

$∴$ $N_{A} I_{A} = 4 N_{B} I_{B}$

Q 13 :

Given below are two statements:

Statement I: If the number of turns in the coil of a moving coil galvanometer is doubled, then the current sensitivity becomes double.

Statement II: Increasing current sensitivity of a moving coil galvanometer by only increasing the number of turns in the coil will also increase its voltage sensitivity in the same ratio.

In the light of the above statements, choose the correct answer from the options given below: [2023]

Statement I is false but Statement II is true
Both Statement I and Statement II are true
Both Statement I and Statement II are false
Statement I is true but Statement II is false

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4

(4)

$For a moving coil galvanometer$

$B i N A = k θ$

$θ = (\frac{B N A}{k}) i; current sensitivity = \frac{B N A}{k}$

$So, if N is doubled then current sensitivity is doubled.$

$Voltage sensitivity$

$B \frac{V}{R} N A = k θ$

$V = \frac{B N A}{R k} θ as N is doubled R is also doubled.$

$So, no change in voltage sensitivity.$

$Hence, option (4) is right.$

Q 14 :

The current sensitivity of moving coil galvanometer is increased by 25%. This increase is achieved only by changing the number of turns of coils and area of cross section of the wire while keeping the resistance of galvanometer coil constant. The percentage change in the voltage sensitivity will be: [2023]

+25%
−50%
Zero
−25%

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(1)

$I_{s} = \frac{N B A}{C} & V_{s} = \frac{N B A}{C G}$

$\Rightarrow V_{s} = \frac{I_{s}}{G}$

$If G (galvanometer resistance) is constant, then V_{s} \propto I_{s}$

$So percentage change in V_{s} is also 25 % .$

Topic Question Set

Q 12 :

The magnetic moments associated with two closely wound circular coils A and B of radius $r_{A} = 10$ cm and $r_{B} = 20$ cm respectively are equal if (where $N_{A}, I_{A}$ and $N_{B}, I_{B}$ are number of turns and current of A and B respectively). [2023]

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