First slide

Magnetic field due to a current carrying circular ring

Question

A straight conductor carrying current i splits into two parts as shown in the figure. The radius of the circular loop is R. The total magnetic field at the centre P of the loop is,

Moderate

A

$\frac{μ_{0} i}{2 R}, inward$
B

Zero
C

$3 μ_{0} i / 32 R, outward$
D

$3 μ_{0} i / 32 R, inward$

Solution

Net magnetic field at point 'P'
$B_{net} = \vec{B_{1}} + \vec{B_{2}}$
Here $\vec{B_{1}} and \vec{B_{2}}$ are equal in magnitude and opposite in direction.
Hence, $B_{n e t} = B_{1} - B_{2}$
$i_{1} = i (\frac{θ}{2 π}) \Rightarrow B_{1} = \frac{μ_{0} i_{1}}{2 R} (\frac{2 π - θ}{2 π})$
$i_{2} = i (\frac{2 π - θ}{2 π}) \Rightarrow B_{2} = \frac{μ_{0} i_{2}}{2 R} (\frac{θ}{2 π})$
$B_{net} = B_{1} - B_{2} = 0$

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