First slide

Magnetic field due to current element

Question

A pair of stationary and infinitely long bent wires is placed in he xy plane as shown in figure. Each wire carries current of 10 A. Segments L and M are along the x-axis. Segments P and Q are parallel to the y-axis such that $O S = O R = 0.02 m$ . Find the magnitude and direction of the magnetic induction at origin ‘O’

Moderate

A

$10^{- 4} T$
B

$10^{- 6} T$
C

$10^{- 8} T$
D

$10^{- 2} T$

Solution

As point ‘O’ is along the length of segments ‘L’ and M, so the field at ‘O’ due to these segments will
be zero. Also point ‘O’ is near one end of along wire.
The resultant field at ‘O’. $B_{R} = B_{P} + B_{Q}$ . This field due to both P and Q will be into the plane of paper, hence will add up.
$\Rightarrow B_{R} = \frac{μ_{0}}{4 π} \frac{I}{R O} + \frac{μ_{0}}{4 π} \frac{I}{S O}$
But $R O = S O = 0.02 M$
Hence, $B_{R} = 2 \times \frac{μ_{0}}{4 π} \times \frac{10}{0.02} = 2 \times 10^{- 7} \times \frac{10}{0.02} = 10^{- 4} T$ .

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App