First slide

Magnetic field due to a straight current carrying wire

Question

An infinitely long conductor PQR is bent to form a right angle as shown. A current f flows through PQR. The magnetic field due to this current at the point M is H₁. Now another infinitely long straight conductor QS is connected at Q so that the current is I/2 in QR as well as in QS. The current in PQ remains unchanged. The magnetic field at M is now H₂. The ratio H₁/H₂ is given by

Moderate

A

$\frac{1}{2}$
B

1
C

$\frac{2}{3}$
D

2

Solution

Magnetic field at any point lying on the axis of a current carrying straight conductor is zero.
Here H₁ = Magnetic field at M due to current in PQ.= $\frac{μ_{0} I}{4 πr}$
H₂ = Magnetic field at M due to QR
+ magnetic field at M due to QS
+ magnetic field at M due to PQ
= 0 + $\frac{μ_{0} I / 2}{4 πr}$ + $\frac{μ_{0} I}{4 π r}$ = $\frac{3}{2} \frac{μ_{0} I}{4 π r}$ . Therefore $\frac{H_{1}}{H_{2}} = \frac{2}{3}$

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