Question

The magnetic induction at a point $1\mathop A\limits^0$ away from a proton measured along its axis of spin is (magnetic moment of the proton is $1.4 \times {10^{ - 26}}A{m^2}$ )
A. $0.28mT$
B. $28mT$
C. $0.028mT$
D. $2.8mT$

Answer 1

Magnetic induction is the possible electromotive force generated in a conductor (or at a point) in a changing magnetic field. You can calculate the magnetic induction by using the equation $B = \dfrac{{{\mu _0}}}{{4\pi }} \times \dfrac{{2m}}{{{r^3}}}$ .

Complete answer:
In the problem we are given that the magnetic moment of the proton is $1.4 \times {10^{ - 26}}A{m^2}$ and using this we have to calculate the value of magnetic induction at a point $1\mathop A\limits^0$ away from a proton measured along its axis of spin.
Before moving on to the mathematical calculations let’s discuss what magnetic moment and magnetic i is.
The magnetic moment is the magnetic strength of a magnetic or any other object that behaves as a magnet and produces magnetic fields (some examples are electric currents, moving elementary particles such as electrons and protons). A magnetic moment can also be called a magnetic dipole moment. The dipole is the magnetic north and south pole separated by a very small distance.
Magnetic induction is the possible electromotive force generated in a conductor in a changing magnetic field.
The equation for magnetic field induction is
$B = \dfrac{{{\mu _0}}}{{4\pi }} \times \dfrac{{2m}}{{{r^3}}}$
Here, $B =$ The magnetic induction due to the spinning proton
${u_0} =$ Magnetic susceptibility of free space $= 4\pi \times {10^{ - 7}}N/{A^2}$
$m =$ The magnetic moment or magnetic dipole moment of the proton
$r =$ The distance of the point from the axis of rotation of the proton
Given in the problem, $r = 1\mathop A\limits^0 = {10^{ - 10}}m$
$m = 1.4 \times {10^{ - 26}}A{m^{ - 2}}$
So, for the given problem
$B = \dfrac{{4\pi \times {{10}^{ - 7}}}}{{{{10}^{ - 7}}}} \times \dfrac{{2 \times 1.4 \times {{10}^{ - 26}}}}{{{{\left( {{{10}^{ - 10}}} \right)}^3}}}$
$B = \dfrac{{4\pi \times {{10}^{ - 7}}}}{{{{10}^{ - 7}}}} \times \dfrac{{2 \times 1.4 \times {{10}^{ - 26}}}}{{{{10}^{30}}}}$
$B = 2.8 \times {10^{ - 3}} = 2.8mT$
Hence, the magnetic induction at a point $1\mathop A\limits^0$ away from a proton measured along its axis of spin is $2.8mT$ (magnetic moment of the proton is $1.4 \times {10^{ - 26}}A{m^2}$ )

So, the correct answer is “Option D”.

Note:
In the situation given to us the spinning proton will produce a north and south pole, so it will behave like a magnetic dipole and have a certain value of magnetic moment (magnetic dipole moment). The magnetic moment is very similar to but electric dipole is made up of two equal and opposite charges instead of north and south poles.