Question #88dab

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Answer 1 · 2017-03-24T14:14:51

The magnetic field wil halved.

At the center of the loop, the magnetic field is given by

$B = \frac{μ_{o} I}{2 r}$ $B=(mu_o I)/(2r)$

Where

the radius $= r$ $=r$

and the diameter is $d = 2 r$ $d=2r$

So,

$B = \frac{μ_{o} I}{d}$ $B=(mu_o I)/(d)$

If we double $d$ $d$ , that is

$D = 2 d$ $D=2d$

Then,

$B_{1} = \frac{μ_{o} I}{D} = \frac{μ_{o} I}{2 d} = \frac{B}{2}$ $B_1=(mu_o I)/(D)=(mu_o I)/(2d)=B/2$

Therefore,

The magnetic field wil be half the value