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hos 5. A long solid right circular cylinder of radius R carries a current I, which is uniformly distributed. Find the magneti

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5. Inside the cyllinder, Ampere's Law can be written as

\oint \bar{B}\cdot \bar{dl}=\mu _{0}I_{enc}

r< R

B2\pi r=\mu _{0}I\frac{\pi r^{2}}{\pi R^{2}}

B_{in} =\frac{\mu _{0}Ir}{2\pi R^{2}}

And for the outer region

r> R

B2\pi r=\mu _{0}I

B_{out}=\frac{\mu _{0}I}{2\pi r}

6. For points inside the solenoid we can write Faraday's Law as

\oint \bar{E}\cdot \bar{dl}=-\frac{d\Phi _{B}}{dt}

r< R_{1}

E2\pi r=-\pi r^{2}\frac{dB}{dt}

Where the B field inside the solenoid is

B=\mu _{0}nI

Therefore

\frac{dB}{dt}=\mu _{0}n\frac{dI}{dt}=-\mu _{0}nI_{0}\omega \sin \omega t

Replacing back we get

E_{in}= \frac{r\mu _{0}nI_{0}\omega \sin \omega t}{2}

For points outside

r> R_{1}

E2\pi r=-\pi R^{2}_{1}\frac{dB}{dt}

E_{out}=\frac{R^{2}_{1}\mu _{0}nI_{0}\omega \sin \omega t}{2 r}

7. We use the lens equation and the magnification formula:

d_{o}=\frac{fd_{i}}{d_{i}-f}

-\frac{d_{i}}{d_{o}}=3

Combining this two

d_{o}=\frac{f(-3d_{o})}{(-3d_{o})-f}

Solving for do we get

d_{o}=\frac{2}{3}f=\frac{20}{3}cm

8. The formula for diffraction grating is

d\sin \theta =n\lambda

Where d is the slit spacing. The condition for 11 interference fringes within the central maximum is

d\sin 90^{\circ} =11\lambda

d =11\lambda

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