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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Electromagnetic Induction
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Consider a solenoid coil with $N= 10$ turns with radius $12 \times 10^{−2}m$. The solenoid is subjected to a varying magnetic field that changes uniformly from $5 T \rightarrow 2 T$ in an interval of $12 s$. The axis of the solenoid makes an angle of $13 ^{\circ}$ to the magnetic field. If the angle is changed to $67 ^{\circ}$, what would the radius need to be for the emf to remain the same?

$\begin{array}{1 1} 0.1994\;m \\ 0.1894\;m \\ 0.5994 \;m \\ 0.0359\;m\end{array}$

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1 Answer

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  • Electromagnetic induction $\varepsilon = N \large\frac {d\Phi}{dt}$
Electromagnetic induction $\varepsilon = N \large\frac {d\Phi}{dt}$$ = N \large\frac {\Delta B A \cos \theta}{\Delta t }$ $ = NA\cos \theta \large\frac{B_2 - B_1}{\Delta T}$
Given $N = 10, \theta = 13 ^{\circ}, r = 12 \times 10^{-2} m, B_1 = 2 T, B_2 = 5T, \Delta t = 12 s$
$\varepsilon = 10 \times \pi \times (12 \times 10^{-2})^2 \times \cos 13 \times \large\frac{3-5}{12}$
$\varepsilon \approx -0.0734 V$
Now, If the angle is changed to $67 ^{\circ}$, we need to caclulate the radius, assuming the emf stays the same.
$\Rightarrow -0.0734 = 10 \times \pi \times (r)^2 \times \cos 67 \times \large\frac{3-5}{12}$
$\Rightarrow -0.0734 = -2.0448 r^2$
$\Rightarrow r = \sqrt 0.03589$
$\Rightarrow r \approx 0.1894 m$
answered Mar 12, 2014 by balaji.thirumalai
 

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