Find the magnetic field at point O. | vedclass

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Question

(A) The magnetic field at the center of a circular arc of radius $R$ subtending an angle $	heta$ at the center is given by $B = \frac{\mu_0 I 	heta}

Vedclass · Accepted Answer

$\pi 	imes 10^{-5} \, T$

Vedclass · Answer

(A) The magnetic field at the center of a circular arc of radius $R$ subtending an angle $	heta$ at the center is given by $B = \frac{\mu_0 I 	heta}{4 \pi R}$.Here,the arc is a quarter circle,so $	heta = \frac{\pi}{2}$ radians.The straight wire segments are directed towards or away from point $O$,so the magnetic field due to them at point $O$ is zero.Thus,the total magnetic field at $O$ is due to the quarter-circular arc only:$B = \frac{1}{4} 	imes \frac{\mu_0 I}{2 R} = \frac{\mu_0 I}{8 R}$.Given: $I = 10 \, A$,$R = 5 \, cm = 5 	imes 10^{-2} \, m$,and $\mu_0 = 4 \pi 	imes 10^{-7} \, T \cdot m/A$.Substituting the values:$B = \frac{4 \pi 	imes 10^{-7} 	imes 10}{8 	imes 5 	imes 10^{-2}} = \frac{40 \pi 	imes 10^{-7}}{40 	imes 10^{-2}} = \pi 	imes 10^{-5} \, T$.

Find the magnetic field at point $O$.

Similar Questions

Assertion : The magnetic field at the centre of the circular coil in the following figure due to the currents $I_1$ and $I_2$ is zero.
Reason : $I_1 = I_2$ implies that the fields due to the current $I_1$ and $I_2$ will be balanced.

$A$ tightly wound coil of $200$ turns and of radius $20 \ cm$ carries a current of $5 \ A$. The magnetic field at the centre of the coil is:

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