Question

A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per meter, calculate (a) H, (b) B and (c) the magnetic current ${I_m}$.

Answer 1

We have heard the term of solenoid in Lenz law where the bar magnet moves inside the solenoid to produce electric current. A long straight wire which is wound up to produce a strong magnetic field inside itself is known as a solenoid.

Complete step by step solution:

Step 1: Calculate the magnetic field:
$B = {\mu _o}{\mu _r}nI$;
Put the given values in the above equation and solve:
$B = 4\pi \times {10^{ - 7}} \times 1000 \times 2 \times 400$;
$B = 4\pi \times {10^{ - 7}} \times 800000$;
The magnetic field comes out to be:
$B = 4\pi \times {10^{ - 2}}T$;

Step 2:
Calculate H:
Formula for Magnetic Field Strength H.
$B = \mu H$;
But $B = {\mu _o}{\mu _r}nI$, so:
$B = \mu H = {\mu _o}{\mu _r}nI$;
Cancel out the common terms:
$H = nI$;
$H = 1000 \times 2$;
$H = 2000A/m$;

Step 3:
Find the magnetic current ${I_m}$.

We have:
$B = {\mu _o}{\mu _r}(I + {I_m})$;
Put the values in the above equation:
$4\pi \times {10^{ - 2}} = 400 \times 4\pi \times {10^{ - 7}}(2 + {I_m})$;
Solve the above equation;
$\dfrac{{{{10}^{ - 2}}}}{{4 \times {{10}^{ - 5}}}} = (2 + {I_m})$;
$\dfrac{{{{10}^3}}}{4} - 2 = {I_m}$;
Simplify the equation:
$\dfrac{{242}}{4} = {I_m}$;
The magnetic current is:
${I_m} = 60.5$;

Final Answer:(a) H = 2000A/m, (b) $B = 4\pi \times {10^{ - 2}}T$ and (c) the magnetic current ${I_m} = 60.5A$

Note : Here we have to find the magnetic field, magnetic current and magnetic field strength. Go step by step, first find out the magnetic field then find out the magnetic current and then magnetic field strength.