(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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Class 12

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Electromagnetic Induction

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Motional EMI (Induced Parameter)

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$A$ square loop $PQRS$ having $10$ turns, area $3.6 \times 10^{-3} \, m^2$ and resistance $100 \, \Omega$ is slowly and uniformly being pulled out of a uniform magnetic field of magnitude $B=0.5 \, T$ as shown. Work done in pulling the loop out of the field in $1.0 \, s$ is . . . . . $\times 10^{-6} \, J$.

DifficultJEE MAIN 2024

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$A$ small conducting loop of radius $a$ and resistance $r$ is pulled with velocity $v$ perpendicular to a long straight conductor carrying a current $i_0$. If a constant power $P$ is dissipated in the loop,find the variation of velocity $v$ of the loop as a function of $x$. Given that $x >> a$.

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$A$ conducting rod with resistance $r$ per unit length is moving inside a vertical magnetic field $\overrightarrow B$ at speed $v$ on two horizontal parallel ideal conductor rails. The ends of the rails are connected to a resistor $R$. The separation between the rails is $d$. The rod maintains a tilted angle $\theta$ to the rails. Find the external force required to keep the rod moving.

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Suppose a long rectangular loop of width $w$ is moving along the $x$-direction with its left arm in a magnetic field perpendicular to the plane of the loop (see figure). The resistance of the loop is zero and it has an inductance $L$. At time $t=0$,its left arm passes the origin,$O$. If for $t \geq 0$,the current in the loop is $I$ and the distance of its left arm from the origin is $x$,then $I$ versus $x$ graph will be

KVPY 2020

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The current-carrying wire and the rod $AB$ are in the same plane. The rod moves parallel to the wire with a velocity $v$. Which one of the following statements is true about the induced emf in the rod?

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$A$ small conducting loop of radius $a$ and resistance $r$ is pulled with velocity $v$ perpendicular to a long straight conductor carrying a current $i_0$. If a constant power $P$ is dissipated in the loop,find the variation of velocity $v$ of the loop as a function of $x$. Given that $x >> a$.

The current-carrying wire and the rod $AB$ are in the same plane. The rod moves parallel to the wire with a velocity $v$. Which one of the following statements is true about the induced emf in the rod?

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