Question

The magnetic field at the centre of a circular coil of radius r is $\pi$ times that due to a long straight wire at a distance r from it, for equal currents. Figure here shows three cases : in all cases the circular part has radius r and straight ones are infinitely long. For same current the B field at the centre P in cases 1, 2, 3 have the ratio

• A. $(-\frac{\pi }{2}):\left(\frac{\pi }{2}\right):(\frac{3\pi }{4}-\frac{1}{2})$
  
• B. $(-\frac{\pi }{2}+1):(\frac{\pi }{2}+1):(\frac{3\pi }{4}-\frac{1}{2})$
  
• C. $-\frac{\pi }{2}:\frac{\pi }{2}:3\frac{\pi }{4}$
  
• D. $(-\frac{\pi }{2}-1):(\frac{\pi }{2}-\frac{1}{4}):(\frac{3\pi }{4}-\frac{1}{2})$

Answer 1

The correct option is A $(-\frac{\pi }{2}):\left(\frac{\pi }{2}\right):(\frac{3\pi }{4}-\frac{1}{2})$




Case 1: $B_A=\frac{{\mu }_0}{4\pi }.\frac{i}{r}\otimes$
$B_B=\frac{{\mu }_0}{4\pi }.\frac{\pi i}{r}\odot$
$B_C=\frac{{\mu }_0}{4\pi }.\frac{i}{r}\odot$
So net magnetic field at the centre of case 1
$B_1=B_B+B_C-B_A\Rightarrow B_1=\frac{{\mu }_0}{4\pi }.\frac{\pi i}{r}\odot$ ...(i)


Case 2 : As we discussed before magnetic field at the centre O in this case

The magnetic field due to a straight current carrying conductor at any point along its length is zero hence,
$B_2=\frac{{\mu }_0}{4\pi }.\frac{\pi i}{r}\otimes$ ...(ii)

Case 3: $B_A=0$


$B_B=\frac{{\mu }_0}{4\pi }.\frac{(2\pi -\pi /2)i}{r}\otimes$
$B_B=\frac{{\mu }_0}{4\pi }.\frac{3\pi i}{2r}\otimes$
$B_C=\frac{{\mu }_0}{4\pi }.\frac{i}{r}\odot$

So net magnetic field at the centre of case 3
$B_C=\frac{{\mu }_0}{4\pi }.\frac{i}{r}(\frac{3\pi }{2}-1)\otimes$ ... (iii)
From equation (i), (ii) and (iii)
$B_1:B_2:B_3=\pi \odot :\pi \otimes (\frac{3\pi }{2}-1)\otimes =-\frac{\pi }{2}:\frac{\pi }{2}:(\frac{3\pi }{4}-\frac{1}{2})$