The magnetic field inside a 4.0 cm diameter superconducting solenoid varies sinusoidally between $8.0T$and$12.0T$at a frequency of$10Hz$.$12.0T$

a. What is the maximum electric field strength at a point $1.5cm$ from the solenoid axis?

b. What is the value of localid="1650012451022" $B$at the instant localid="1650084655519" $E$reaches its maximum value?

We have been given that the magnetic field inside a $4.0cm$diameter solenoid varies sinusoidally between $8.0T$and $12.0T$at a frequency of $10Hz$.

We need to calculate the maximum electric field strength at a point$1.5cm$from the solenoid axis.

The solenoid is the current carrying wire. The electric field circles around inside the solenoid.

So at a distance $r<R$,

$\int E.ds=E\left(2\mathrm{\pi r}\right)$and area$A=\pi r^2$.

When magnetic field varies sinusoidally the expression for magnetic field islocalid="1650011266998" $B\left(t\right)=B_c+B_o\sin \left(2\mathrm{\pi ft}\right)$

Substitute $10.0T$forlocalid="1650012464363" $B_c$,$2.0T$for$B_o$and $10Hz$for localid="1650084726782" $f$

$$\begin{gathered} B\left(t\right)=(10.0T)+(2.0T)\sin \left(2\mathrm{\pi }\right(10\mathrm{Hz}\left)\mathrm{t}\right) \\ =(10.0T)+(2.0T)\sin \left(20\mathrm{\pi t}\right) \end{gathered}$$

differentiate $B\left(t\right)$

$\frac{d\left(B\right(t\left)\right)}{dt}=\left(40\mathrm{\pi T}\right)\cos \left(20\mathrm{\pi T}\right)$

Substitute${\mathrm{\pi r}}^2$forlocalid="1650084219494" $A$,$E.\left(2\mathrm{\pi r}\right)$for $\int E.ds$,and $\left(40\mathrm{\pi T}\right)\cos \left(20\mathrm{\pi t}\right)$for $\raisebox{1ex}{$d\left(B\right(t\left)\right)$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.$

$$\begin{gathered} E.\left(2\mathrm{\pi r}\right)=-\left(\right(40\mathrm{\pi T}\left)\cos \right(20\mathrm{\pi t}\left)\right)\left({\mathrm{\pi r}}^2\right) \\ \mathrm{E}=-\frac{\left(\right(40\mathrm{\pi T}\left)\cos \right(20\mathrm{\pi t}\left)\right)\left(\mathrm{\pi r}\right)}{\left(2\mathrm{\pi }\right)} \end{gathered}$$

The electric field will be maximum when $\cos \left(20\mathrm{\pi t}\right)=1$

Substitute$\cos \left(20\mathrm{\pi t}\right)=1$rewrite the above expression in terms of $E_{max}$

$E_{max}=\frac{\left(\right(40\mathrm{\pi t}\left)1\right).\left(\mathrm{\pi r}\right)}{\left(2\mathrm{\pi }\right)}$

Substitute,$1.5cm$for localid="1650084740352" $r$to find $E_{max}$,

$$\begin{gathered} E_{max}=-\frac{\left(40\mathrm{\pi t}\right)\left(1.5\mathrm{cm}\left({\displaystyle \frac{1\mathrm{m}}{{10}^2\mathrm{cm}}}\right)\right)}{\left(2\mathrm{\pi }\right)} \\ =0.3V/m \end{gathered}$$

We have been given that the magnetic field inside a $4.0cm$diameter superconducting solenoid varies sinusoidally between $8.0T$and $12.0T$at a frequency of $10Hz$.

We need to find the value of $B$at the instant $E$reaches its maximum value.

The electric field reaches maximum when,
$20\mathrm{\pi t}=\mathrm{\pi },3\mathrm{\pi },5\mathrm{\pi },..$

where sine function goes to zero,

So when electric field reaches maximum

$B=B_c+B_0\sin \left(\pi \right)$
$=B_c$
$=10T$