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Question:
An infinitely long conductor PQR is bent to form a right angle as shown in figure. A current I flows through PQR. The magnetic field due to this current at the point M is $H_1$. Now, another infinitely long straight conductor QS is connected at Q, so that current is I/2 in QR as well as in QS, the current in PQ remaining unchanged. The magnetic field at M is now $H_2.$ The ratio $H_1 /H_2$ is given by
An infinitely long conductor PQR is bent to form a right angle as shown in figure. A current I flows through PQR. The magnetic field due to this current at the point M is $H_1$. Now, another infinitely long straight conductor QS is connected at Q, so that current is I/2 in QR as well as in QS, the current in PQ remaining unchanged. The magnetic field at M is now $H_2.$ The ratio $H_1 /H_2$ is given by
Updated On: Jun 14, 2022
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The Correct Option is C
Solution and Explanation
$H_1$ = Magnetic field at M due to PQ + Magnetic field at M due to QR
But magnetic field at M due to QR = 0
$\therefore$ Magnetic field at M due to PQ (or due to current 1 in PQ)$=H_1$
Now $H_2$ = Magnetic field at M due to PQ (current I)
$\, \, \, \, \, \, \, \, \, \, $+ magnetic field at M due to QS (current 1/2)
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, $+ magnetic field at M due to QR
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =H_1+\frac{H_1}{2}+0=\frac{3}{2}H_1$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \frac{H_1}{H_2}=\frac{2}{3}$
NOTE Magnetic field at any point lying on the current carrying straight conductor is zero.
But magnetic field at M due to QR = 0
$\therefore$ Magnetic field at M due to PQ (or due to current 1 in PQ)$=H_1$
Now $H_2$ = Magnetic field at M due to PQ (current I)
$\, \, \, \, \, \, \, \, \, \, $+ magnetic field at M due to QS (current 1/2)
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, $+ magnetic field at M due to QR
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =H_1+\frac{H_1}{2}+0=\frac{3}{2}H_1$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \frac{H_1}{H_2}=\frac{2}{3}$
NOTE Magnetic field at any point lying on the current carrying straight conductor is zero.
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Concepts Used:
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Magnetic Field:
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F = q [ E (r) + v × B (r)] ≡ EElectric +Fmagnetic
This force was based on the extensive experiments of Ampere and others. It is called the Lorentz force.
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