$\left(\text{Take } \pi = \frac{22}{7}\right)$
Physics Solution: Induced EMF in Rotating Loop
This problem requires calculating the radius of a conducting circular loop using the parameters of its rotation in a magnetic field and the resulting induced electromotive force (EMF).
Given Parameters
- Angular speed, $\omega = 100 \text{ rad/s}$
- Magnetic field strength, $B = 0.5 \text{ T}$
- Angle of rotation from horizontal, $\theta = 30^\circ$
- Induced EMF, $\mathcal{E} = 15.4 \text{ mV} = 15.4 \times 10^{-3} \text{ V}$
- Value of pi, $\pi = \frac{22}{7}$
Induced EMF Formula
The EMF induced in a conducting loop rotating in a uniform magnetic field is given by:
$ \mathcal{E} = B A \omega \sin(\theta) $
Here, $A$ represents the area of the loop, which for a circular loop is $A = \pi r^2$, where $r$ is the radius.
Substituting the area formula, the EMF equation becomes:
$ \mathcal{E} = B (\pi r^2) \omega \sin(\theta) $
Calculating Loop Radius
To find the radius $r$, we first rearrange the formula to solve for $r^2$:
$ r^2 = \frac{\mathcal{E}}{B \pi \omega \sin(\theta)} $
Substitute the provided values into the equation:
$ r^2 = \frac{15.4 \times 10^{-3} \text{ V}}{(0.5 \text{ T}) \times (\frac{22}{7}) \times (100 \text{ rad/s}) \times \sin(30^\circ)} $
Using $\sin(30^\circ) = 0.5 = \frac{1}{2}$:
$ r^2 = \frac{15.4 \times 10^{-3}}{0.5 \times \frac{22}{7} \times 100 \times 0.5} $
Simplify the denominator:
$ 0.5 \times \frac{22}{7} \times 100 \times 0.5 = \frac{1}{2} \times \frac{22}{7} \times 50 = \frac{11}{7} \times 50 = \frac{550}{7} $
Now calculate $r^2$:
$ r^2 = \frac{15.4 \times 10^{-3}}{\frac{550}{7}} = \frac{15.4 \times 10^{-3} \times 7}{550} $
$ r^2 = \frac{107.8 \times 10^{-3}}{550} = 0.196 \times 10^{-3} \text{ m}^2 $
Convert this to a more convenient form for square root calculation:
$ r^2 = 196 \times 10^{-6} \text{ m}^2 $
Take the square root to find the radius $r$:
$ r = \sqrt{196 \times 10^{-6} \text{ m}^2} $
$ r = 14 \times 10^{-3} \text{ m} $
Final Result
The calculated radius is $14 \times 10^{-3}$ meters. Converting this to millimeters:
$ r = 14 \text{ mm} $
The radius of the loop is 14 mm.
Similar Questions
- The electric field of a plane electromagnetic wave, travelling in an unknown non-magnetic medium is given by,
$E_y = 20 \sin(3 \times 10^6 x - 4.5 \times 10^{14} t) \text{ V/m}$
(where x, t and other values have S.I. units). The dielectric constant of the medium is _________.
(speed of light in free space is $3 \times 10^8 \text{ m/s}$) Figure shows the circuit that contains three resistances ($9 \, \Omega$ each) and two inductors (4 mH each). The reading of ammeter at the moment switch K is turned ON, is _________ A.
- A capacitor $P$ with capacitance $10 \times 10^{-6} \text{ F}$ is fully charged with a potential difference of 6.0 V and disconnected from the battery. The charged capacitor $P$ is connected across another capacitor $Q$ with capacitance $20 \times 10^{-6} \text{ F}$. The charge on capacitor $Q$ when equilibrium is established will be $\alpha \times 10^{-5} \text{ C}$ (assume capacitor $Q$ does not have any charge initially), the value of $\alpha$ is _________.
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(electron concentration $= 5 \times 10^{28} \text{/m}^3$ and electron charge = $1.6 \times 10^{-19} \text{ C}$) - There are three co-centric conducting spherical shells $A$, $B$ and $C$ of radii $a$, $b$ and $c$ respectively ($c > b > a$) and they are charged with charge $q_1$, $q_2$ and $q_3$ respectively. The potentials of the spheres $A$, $B$ and $C$ respectively, are :
For the series $LCR$ circuit connected with 220 V, 50 Hz a.c source as shown in the figure, the power factor is $\frac{\alpha}{10}$. The value of $\alpha$ is ______.
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(permittivity of vacuum is $\epsilon_0$) Two resistors $2\, \Omega$ and $3\, \Omega$ are connected in the gaps of bridge as shown in figure. The null point is obtained with the contact of jockey at some point on wire $XY$. When an unknown resistor is connected in parallel with $3\, \Omega$ resistor, the null point is shifted by 22.5 cm toward $Y$. The resistance of unknown resistor is ______ $\Omega$.
Match the LIST-I with LIST-II
List-I: List-II: A. Radio-wave I. is produced by Magnetron valve B. Micro-wave II. due to change in the vibrational modes of atoms C. Infrared-wave III. due to inner shell electrons moving from higher energy level to lower energy level D. X-ray IV. due to rapid acceleration of electrons Choose the correct answer from the options given below:
Important Questions from Electricity and Magnetism
- The electric field of a plane electromagnetic wave, travelling in an unknown non-magnetic medium is given by,
$E_y = 20 \sin(3 \times 10^6 x - 4.5 \times 10^{14} t) \text{ V/m}$
(where x, t and other values have S.I. units). The dielectric constant of the medium is _________.
(speed of light in free space is $3 \times 10^8 \text{ m/s}$) Figure shows the circuit that contains three resistances ($9 \, \Omega$ each) and two inductors (4 mH each). The reading of ammeter at the moment switch K is turned ON, is _________ A.
- A capacitor $P$ with capacitance $10 \times 10^{-6} \text{ F}$ is fully charged with a potential difference of 6.0 V and disconnected from the battery. The charged capacitor $P$ is connected across another capacitor $Q$ with capacitance $20 \times 10^{-6} \text{ F}$. The charge on capacitor $Q$ when equilibrium is established will be $\alpha \times 10^{-5} \text{ C}$ (assume capacitor $Q$ does not have any charge initially), the value of $\alpha$ is _________.
- A cylindrical conductor of length 2 m and area of cross-section $0.2 \text{ mm}^2$ carries an electric current of 1.6 A when its ends are connected to a 2 V battery. Mobility of electrons in the conductor is $\alpha \times 10^{-3} \text{ m}^2\text{/V.s}$. The value of $\alpha$ is :
(electron concentration $= 5 \times 10^{28} \text{/m}^3$ and electron charge = $1.6 \times 10^{-19} \text{ C}$) - There are three co-centric conducting spherical shells $A$, $B$ and $C$ of radii $a$, $b$ and $c$ respectively ($c > b > a$) and they are charged with charge $q_1$, $q_2$ and $q_3$ respectively. The potentials of the spheres $A$, $B$ and $C$ respectively, are :
