Question

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

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

b. What is the value of B at the instant E reaches its maximum value?

Short answer

(a) The electric field at $1.5cm$above the axis of solenoid is $0.3V/m$

(b) The value of$B$the instant$E$reaches its maximum is${10}^T$

Step-by-step solution

Part(a) Step 1: Explanation

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

So at a distance $r<R$.

$\int E·ds=E\left(2\pi r\right)$

And area $A=\pi r^2$

Part(a) Step 2:

When magnetic field varies sinusoidally the expression for magnetic field is.

$B\left(t\right)=B_c+B_0\sin \left(2\pi ft\right)$

Substitute $10.0\mathrm{T}$for $B_c$, $2.0\mathrm{T}$for $B_0$and $10Hz$for $f$.

role="math" localid="1649862479221" $B\left(t\right)=(10.0T)+(2.0T)\sin \left(2\mathrm{\pi }\right(10\mathrm{Hz}\left)\mathrm{t}\right)$

role="math" localid="1649862488387" $=(10.0\mathrm{T})+(2.0\mathrm{T})\sin \left(20\pi t\right)$

Differentiate $B\left(t\right)$

role="math" localid="1649862500756" $\frac{d\left(B\right(t\left)\right)}{dt}=\left(40\pi \mathrm{T}\right)\cos \left(20\pi t\right)$

Substitute $\pi r^2$for $A,E·\left(2\pi r\right)$for $\int E·ds$, and role="math" localid="1649862517441" $\left(40\pi \mathrm{T}\right)\cos \left(20\pi t\right)$for $d\left(B\right(t\left)\right)/dt$

role="math" localid="1649862530672" $E·\left(2\pi r\right)=-\left(\right(40\pi \mathrm{T}\left)\cos \right(20\pi t\left)\right)\left(\pi r^2\right)$

role="math" localid="1649862541580" $E=-\frac{\left(\right(40\pi \mathrm{T}\left)\cos \right(20\pi t\left)\right)\left(\pi r\right)}{\left(2\pi \right)}$

The electric field will be maximum when role="math" localid="1649862607430" $\cos \left(20\mathrm{\pi t}\right)=1$

Substitute $\cos \left(20\mathrm{\pi t}\right)=1$(to write the above expression in term of$E_{max}$).

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

Substitute $1.5cm$for $r$to find $E_{max}$.

$E_{\max }=-\frac{\left(40\pi \mathrm{T}\right)\left(1.5\mathrm{cm}\left(\frac{1\mathrm{m}}{{10}^2\mathrm{cm}}\right)\right)}{\left(2\pi \right)}$

$=0.3V/m$

Part(b) Step 1: Explanation

The electric field reaches maximum when,

$22\mathrm{\pi t}=\mathrm{\pi },3\mathrm{\pi },5\mathrm{\pi }$,....

Where sine function goes to zero.

So when electric field reaches maximum,

$B=B_e+B_{o_{}}\sin \left(\mathrm{\pi }\right)$

$=B_e$

$=10T$