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

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

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Similar Questions

$A$ conducting rod of length $L = 0.1 \, m$ is moving with a uniform speed $v = 0.2 \, m/s$ on conducting rails in a magnetic field $B = 0.5 \, T$ as shown. On one side,the end of the rails is connected to a capacitor of capacitance $C = 20 \, \mu F$. Then the charges on the capacitor plates $A$ and $B$ are:

$A$ horizontal straight wire $10 \; m$ long extending from east to west is falling with a speed of $5.0 \; m \, s^{-1}$,at right angles to the horizontal component of the earth's magnetic field,$0.30 \times 10^{-4} \; Wb \, m^{-2}$.
$(a)$ What is the instantaneous value of the $emf$ induced in the wire?
$(b)$ What is the direction of the $emf$?
$(c)$ Which end of the wire is at the higher electrical potential?

$A$ rectangular loop $PQMN$ with a movable arm $PQ$ of length $12 \, cm$ and resistance $2 \, \Omega$ is placed in a uniform magnetic field of $0.1 \, T$ acting perpendicular to the plane of the loop as shown in the figure. The resistance of the arms $MN$, $NP$, and $MQ$ are negligible. The current induced in the loop when arm $PQ$ is moved with a velocity of $20 \, ms^{-1}$ is (in $ \, A$)

$A$ thin wire of length $2\,m$ is placed perpendicular to the $x-y$ plane. It is moved with velocity $\overrightarrow v = (2\hat i + 3\hat j + \hat k)\,m/s$ through a region of magnetic induction $\overrightarrow B = (\hat i + 2\hat j)\,Wb/m^2$. The potential difference induced between the ends of the wire is......$V$.

$A$ simple pendulum with a bob of mass $m$ and a conducting wire of length $L$ swings under gravity through an angle $\theta$. The component of the Earth's magnetic field in the direction perpendicular to the swing is $B$. The maximum e.m.f. induced across the pendulum is ($g=$ acceleration due to gravity).

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