A long thin-walled pipe of radius R carries a | vedclass

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(A) Consider a small element of the pipe subtending an angle $d	heta$ at the center. The current in this element is $dI = \frac{I}{2\pi} d	heta$.The

Vedclass · Accepted Answer

$\frac{{\mu _0}I\sqrt 2 }{4{\pi ^2}R}$

Vedclass · Answer

(A) Consider a small element of the pipe subtending an angle $d	heta$ at the center. The current in this element is $dI = \frac{I}{2\pi} d	heta$.The magnetic field $dB$ at the center due to this long current-carrying strip (acting as an infinite wire) is given by $dB = \frac{\mu_0 (dI)}{2\pi R} = \frac{\mu_0 I}{4\pi^2 R} d	heta$.The direction of $dB$ is perpendicular to the radius vector of the element. For the quarter portion,the angle $	heta$ ranges from $0$ to $\pi/2$. Resolving the field into components:$dB_x = dB \sin	heta$ and $dB_y = -dB \cos	heta$.Integrating these components from $0$ to $\pi/2$:$B_x = \int_0^{\pi/2} \frac{\mu_0 I}{4\pi^2 R} \sin	heta d	heta = \frac{\mu_0 I}{4\pi^2 R} [-\cos	heta]_0^{\pi/2} = \frac{\mu_0 I}{4\pi^2 R}$.$B_y = \int_0^{\pi/2} \frac{\mu_0 I}{4\pi^2 R} \cos	heta d	heta = \frac{\mu_0 I}{4\pi^2 R} [\sin	heta]_0^{\pi/2} = \frac{\mu_0 I}{4\pi^2 R}$.The magnitude of the resultant magnetic field is $B = \sqrt{B_x^2 + B_y^2} = \sqrt{(\frac{\mu_0 I}{4\pi^2 R})^2 + (\frac{\mu_0 I}{4\pi^2 R})^2} = \frac{\mu_0 I \sqrt{2}}{4\pi^2 R}$.

List-$I$	List-$II$
$(A)$ Fleming's left hand rule	$(i)$ Direction of induced current
$(B)$ Right hand thumb rule	(ii) Magnitude and direction of magnetic induction
$(C)$ Biot-Savart law	(iii) Direction of force due to magnetic induction
$(D)$ Fleming's right hand rule	(iv) Direction of magnetic lines due to current

$A$ long thin-walled pipe of radius $R$ carries a current $I$ along its length. The current density is uniform over the circumference of the pipe. The magnetic field at the center of the pipe due to the quarter portion of the pipe shown is:

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