(B) Given:Number of turns in the coil, $N = 30$.Area of the coil, $A = 2.5 \, cm^2 = 2.5 \times 10^{-4} \, m^2$.Total charge flowing through the coil,

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(B) Given:Number of turns in the coil, $N = 30$.Area of the coil, $A = 2.5 \, cm^2 = 2.5 \times 10^{-4} \, m^2$.Total charge flowing through the coil, $Q = 7.5 \times 10^{-3} \, C$.Total resistance of the circuit, $R = 0.3 \, \Omega$.We know that the induced charge $Q$ is given by the formula:$Q = \frac{\Delta \phi}{R} = \frac{N B A}{R}$Rearranging the formula to solve for the magnetic field $B$:$B = \frac{Q R}{N A}$Substituting the given values:$B = \frac{(7.5 \times 10^{-3} \, C) \times (0.3 \, \Omega)}{30 \times (2.5 \times 10^{-4} \, m^2)}$$B = \frac{2.25 \times 10^{-3}}{7.5 \times 10^{-3}}$$B = 0.3 \, T$Thus, the magnitude of the magnetic field is $0.3 \, T$.

Class 12 Physics Electromagnetic Induction Motional EMI (Induced Parameter)

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To measure a magnetic field between the magnetic poles of a loudspeaker, a small coil having $30$ turns and $2.5 \, cm^2$ area is placed perpendicular to the field and removed immediately. If the total charge flown through the coil is $7.5 \times 10^{-3} \, C$ and the total resistance of the wire and galvanometer is $0.3 \, \Omega$, then the magnitude of the magnetic field is

AP EAMCET 2019 All AP EAMCET

A
$0.03 \, T$
B
$0.3 \, T$
C
$3 \, T$
D
$3 \times 10^2 \, T$

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Electromagnetic Induction

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Motional EMI (Induced Parameter)

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$A$ metallic disc of radius $0.3 \ m$ is rotating with a constant angular speed of $60 \ rad \ s^{-1}$ in a plane perpendicular to a uniform magnetic field of $5 \times 10^{-2} \ T$. The emf induced between a point on the rim and the centre of the disc is: (in $V$)

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$A$ copper rod $AB$ of length $l$ is rotated about end $A$ with a constant angular velocity $\omega$. The electric field at a distance $x$ from the axis of rotation is

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$A$ cycle wheel contains $24$ spokes of $0.5 \, m$ length. It is rotated in a horizontal plane with $120 \, \text{revolution/min}$ in the presence of the Earth's magnetic field. If the total magnetic field of the Earth is $10^{-4} \, T$ (given $10^4 \, G = 1 \, T$), then find the dynamic $emf$ induced across the centre and the rim of the wheel (angle of dip is $30^{\circ}$).

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$A$ rectangular loop with a sliding connector of length $10\, cm$ is situated in a uniform magnetic field perpendicular to the plane of the loop. The magnetic induction is $0.1\, T$ and the resistance of the connector is $1\, \Omega$. The sides $AB$ and $CD$ have resistances $2\, \Omega$ and $3\, \Omega$ respectively. Find the current in the connector during its motion with a constant velocity of $1\, m/s$.

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$A$ circular coil of radius $30 \ cm$,having $1$ turn and a resistance of ${\pi ^2} \ \Omega$,is placed in a magnetic field of $10^{-2} \ T$. The coil rotates about its diameter with a speed of $200 \ rpm$ in a direction perpendicular to the magnetic field. The value of the induced $AC$ current in the coil will be . . . . . . $mA$.

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$A$ metallic disc of radius $0.3 \ m$ is rotating with a constant angular speed of $60 \ rad \ s^{-1}$ in a plane perpendicular to a uniform magnetic field of $5 \times 10^{-2} \ T$. The emf induced between a point on the rim and the centre of the disc is: (in $V$)

$A$ copper rod $AB$ of length $l$ is rotated about end $A$ with a constant angular velocity $\omega$. The electric field at a distance $x$ from the axis of rotation is

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