First slide

Magnetic field due to a straight current carrying wire

Question

In fig. what is the magnetic field induction at point o

Moderate

A

$\frac{μ_{0} i}{4 πr}$
B

$\frac{μ_{0} i}{4 r} + \frac{μ_{0} i}{2 πr}$
C

$\frac{μ_{0} i}{4 r} + \frac{μ_{0} i}{4 πr}$
D

$\frac{μ_{0} i}{4 r} - \frac{μ_{0} i}{4 πr}$

Solution

B at O will be due to the following portions :
(i) vertical straight portion. This is zero.
(ii) circular portion. This is given by
$B_{circular} = \frac{1}{2} \frac{μ_{0} i}{2 r} = \frac{μ_{0} i}{4 r}$
(iii) straight horizontal portion. This is given by
$\begin{array}{r} B_{straight} = \frac{μ_{0} i}{4 πr} \\ ∴ B_{Total} = \frac{μ_{0} i}{4 r} + \frac{u_{0} i}{4 πr} \end{array}$

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