$A$ line charge $\lambda$ per unit length is lodged uniformly onto the rim of a wheel of mass $M$ and radius $R$. The wheel has light non-conducting spokes and is free to rotate without friction about its axis. $A$ uniform magnetic field extends over a circular region within the rim. It is given by,
$B = -B_{0} \hat{k}$ for $r \leq a$ (where $a < R$)
$B = 0$ otherwise.
What is the angular velocity of the wheel after the field is suddenly switched off?
$B = -B_{0} \hat{k}$ for $r \leq a$ (where $a < R$)
$B = 0$ otherwise.
What is the angular velocity of the wheel after the field is suddenly switched off?
(D) When the magnetic field is switched off,a changing magnetic flux induces an electric field. According to Faraday's law,the induced electromotive force $(EMF)$ is given by $\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt}$.
For a circular path of radius $r \leq a$,the magnetic flux is $\Phi_B = B \cdot \pi r^2$.
The induced electric field $E$ at radius $r$ is $E(2\pi r) = \frac{d}{dt}(B_0 \pi r^2) = \pi r^2 \frac{dB_0}{dt}$.
Thus,$E = \frac{r}{2} \frac{dB_0}{dt}$.
The torque $\tau$ acting on the rim (where charge $Q = \lambda (2\pi R)$ is located) is $\tau = r_f \times F = R(qE)$.
Since the charge is on the rim,$r=R$ for the force calculation,but the induced field is inside. The force on the rim charge $dq$ is $dF = dq E = (\lambda dl) E$.
The total torque is $\tau = \int R E dq = R E (\lambda 2\pi R) = 2\pi R^2 \lambda E$.
Substituting $E$ at $r=a$ (the boundary of the field),the impulse is $\int \tau dt = \Delta L = I\omega$.
$I = MR^2$.
$\int (2\pi R^2 \lambda) \frac{a}{2} \frac{dB_0}{dt} dt = MR^2 \omega$.
$\pi R^2 \lambda a B_0 = MR^2 \omega$.
$\omega = \frac{\pi a^2 \lambda B_0}{MR}$.
For a circular path of radius $r \leq a$,the magnetic flux is $\Phi_B = B \cdot \pi r^2$.
The induced electric field $E$ at radius $r$ is $E(2\pi r) = \frac{d}{dt}(B_0 \pi r^2) = \pi r^2 \frac{dB_0}{dt}$.
Thus,$E = \frac{r}{2} \frac{dB_0}{dt}$.
The torque $\tau$ acting on the rim (where charge $Q = \lambda (2\pi R)$ is located) is $\tau = r_f \times F = R(qE)$.
Since the charge is on the rim,$r=R$ for the force calculation,but the induced field is inside. The force on the rim charge $dq$ is $dF = dq E = (\lambda dl) E$.
The total torque is $\tau = \int R E dq = R E (\lambda 2\pi R) = 2\pi R^2 \lambda E$.
Substituting $E$ at $r=a$ (the boundary of the field),the impulse is $\int \tau dt = \Delta L = I\omega$.
$I = MR^2$.
$\int (2\pi R^2 \lambda) \frac{a}{2} \frac{dB_0}{dt} dt = MR^2 \omega$.
$\pi R^2 \lambda a B_0 = MR^2 \omega$.
$\omega = \frac{\pi a^2 \lambda B_0}{MR}$.
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Similar Questions
$A$ point charge $Q$ is moving in a circular orbit of radius $R$ in the $x$-$y$ plane with an angular velocity $\omega$. This can be considered as equivalent to a loop carrying a steady current $I = \frac{Q\omega}{2\pi}$. $A$ uniform magnetic field along the positive $z$-axis is now switched on,which increases at a constant rate from $0$ to $B$ in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around a closed loop. It is known that,for an orbiting charge,the magnetic dipole moment is proportional to the angular momentum with a proportionality constant $\gamma$.
$1.$ The magnitude of the induced electric field in the orbit at any instant of time during the time interval of the magnetic field change is:
$(A)$ $\frac{BR}{4}$ $(B)$ $\frac{BR}{2}$ $(C)$ $BR$ $(D)$ $2BR$
$2.$ The change in the magnetic dipole moment associated with the orbit,at the end of the time interval of the magnetic field change,is:
$(A)$ $-\gamma BQR^2$ $(B)$ $-\gamma \frac{BQR^2}{2}$ $(C)$ $\gamma \frac{BQR^2}{2}$ $(D)$ $\gamma BQR^2$
Give the answer for question $1$ and $2$.
$1.$ The magnitude of the induced electric field in the orbit at any instant of time during the time interval of the magnetic field change is:
$(A)$ $\frac{BR}{4}$ $(B)$ $\frac{BR}{2}$ $(C)$ $BR$ $(D)$ $2BR$
$2.$ The change in the magnetic dipole moment associated with the orbit,at the end of the time interval of the magnetic field change,is:
$(A)$ $-\gamma BQR^2$ $(B)$ $-\gamma \frac{BQR^2}{2}$ $(C)$ $\gamma \frac{BQR^2}{2}$ $(D)$ $\gamma BQR^2$
Give the answer for question $1$ and $2$.
$A$ time-varying magnetic field $B(t)$ exists in a circular region of radius '$a$' and is directed into the plane of the paper,as shown in the figure. The magnitude of the induced electric field at a point '$P$' at a distance '$r$' $(r > a)$ from the center of the circular region is:
Medium
View SolutionWhich conclusion can we obtain from the fact that an $emf$ is induced in a stationary conductor placed in a time-varying magnetic field? Discuss the characteristics of the induced electric field.
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View SolutionThe figure shows a circular region of radius $R$ in which a uniform magnetic field $B$ exists. The magnetic field is increasing at a rate $\frac{d B}{d t}$. The magnitude of the induced electric field at a distance $r$ from the centre for $r < R$ is ............
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View Solution$A$ circular region of radius $R$ has a uniform magnetic field $B = B_0 + B_0 t(-\hat{k})$. At $t = 0$,what is the acceleration of a charged particle of mass $m$ and charge $q$ placed at a distance $r$ $(r > R)$ from the center?
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