Question

The 10 cm long magnet has pole strength of 12Am. The magnetic field at a distance 20cm from its center:
(A) $2.4 \times {10^{ - 5}}T$
(B) $1.28 \times {10^{ - 5}}T$
(C) $3.8 \times {10^{ - 5}}T$
(D) $3 \times {10^{ - 5}}T$

Answer 1

Hint
Length of the bar magnet is given. We have to calculate the magnetic field at a point which is 20cm away from its center. We will use $B = \dfrac{{2{\mu _0}Md}}{{4\pi {{({d^2} - {l^2})}^2}}}$ to evaluate the value of magnetic field.

Complete step by step answer
Bar magnet: It is an object which shows all properties of magnet. It is rectangular in shape. It experiences a magnetic field up to a certain region. It has 2 poles north as well as south pole.
Pole strength: amount of field lines crossing the either end of the pole. Field lines emerge from the north pole and enter at the south pole from outside.
Magnetic field lines: lines are formed by placing compass needles in nearby regions of magnet. The directions in which the needle shows deflections are joined together to form field lines.
Length of bar magnet be 2l.
 $\Rightarrow 2l = 10cm\,\,(given) $
 $\Rightarrow 2l = 0.1m\, $
 $\Rightarrow l = 0.1/2m = 0.005m $
Pole strength be ‘M’.
 $\Rightarrow M = 12Am$ (given)
Symbol of the magnetic field is ‘B’.
Formula of Magnetic field across axis is $B = \dfrac{{2{\mu _0}Md}}{{4\pi {{({d^2} - {l^2})}^2}}}$ …(1)
In this, ${\mu _0}$ = magnetic susceptibility. It is a unit-less quantity. Its value is $4\pi \times {10^{ - 7}}$ .
Distance from the centre of the magnet at a certain point is ‘d’.
$\Rightarrow d = 20cm\, = 0.2m$
Putting the value of ${\mu _0}$ , M, d and l in equation (1), we get
 $\Rightarrow B = \dfrac{{2 \times 4\pi \times {{10}^{ - 7}} \times 12 \times 0.2}}{{4\pi {{[{{(0.2)}^2} - {{(0.05)}^2}]}^2}}}$
 $\Rightarrow B = 1.28 \times {10^{ - 5}}T$
Since we get the value of the magnetic field as $B = 1.28 \times {10^{ - 5}}T$ ,so the correct option is (E).

Note
Length of magnet is considered as 2l instead of l, our solution will be wrong. We will open squares inside square brackets firstly, then we solve the outer bracket square. If we used values in units of centimeters instead of meters then we don’t get required results.