Question

a circular coil of radius R carries a current I. the magnetic field at its center is B. at what distance from the centre on the axis of the coil, the magnetic field will be B/8

Answer 1

hope this helps you .

Answer 2

Answer:

$\sqrt{3}R$

Explanation:

For these kinds of questions look for a clue word or clue sentence.

Here the clue sentence is on the last line: magnetic field will be B/8

• From our prior knowledge, we know that the Magnetic field is B.
• Here they have given it is B/8
• So you can simply equate B to B/8.

Then,

$B .axis = \frac{B}{8}$

$\frac{\mu0NiR^{2} }{2(R^{2}+x^{2} )^3^/^2 } = \frac{\mu0Ni}{8*2R}$

By cross-multiplying,

${\mu0NiR^{2} }*[ {8*2R}] = {\mu0Ni}* [{2(R^{2}+x^{2} )^3^/^2 }]$

$\frac{\mu0NiR^{2} }{\mu0Ni} = \frac{2(R^{2}+x^{2} )^3^/^2} {8*2R}$

After simplifying and cancelling respective terms,

${R^{2} } = \frac{(R^{2}+x^{2} )^3^/^2} {4*2R}\\\\= {R^{2} } = \frac{(R^{2}+x^{2} )^3^/^2} {8R}\\\\= {R^{2}}* {8R} ={(R^{2}+x^{2} )^3^/^2} \\\\= {8 R^3} ={(R^{2}+x^{2} )^3^/^2}$

This is a simplified expression.

$\hrulefill$

Now keep in mind one general idea.

$x^{2} = 2\\x = \sqrt{2}$

• Do note that here, the power was initially 2, but moves to the other side of the equation and becomes 1/2 which is expressed using $\sqrt{}$ symbol.
• You can also think of it as the power becomes the reciprocal of itself as it moves to the other side of the equation.

$\hrulefill$

Now let us get back to the sum

in the expression, ${8 R^3} ={(R^{2}+x^{2} )^3^/^2}$

By taking 8 as 2*2*2, we can write:

${(2R)^3} ={(R^{2}+x^{2} )^3^/^2}$

The power 3/2 on the right-hand side can be moved to the left. This makes it reciprocate itself and 3/2 becomes 2/3 as told in the example.

$(({2 R)^3})^2^/^3 ={(R^{2}+x^{2} )}$

We also know that, $(a^m)^n = a^mn$

By applying that rule:

3 and 2/3 get multiplied by each other and become 6/3 = 2

$({2 R)^2 ={R^{2}+x^{2} }$

${4R^2 ={R^{2}+x^{2} }$

${4R^2 - {R^{2} =x^{2} }$

${3R^2=x^{2} }$

$\sqrt{3R^2} = x$

±$\sqrt{3}R = x$

Thus your answer is $\sqrt{3}R$