Name: Fundamental Particles
Uploaded: 2019-07-19
Duration: 4 min 20 s
Description: Here is a video on fundamental particles of an atom which elaborates on the list of�sub-atomic particles. Watch this to explore more.

Question

Find the magnetic induction at the point $O$ , if the wire carrying a current $I = 8.0 A$ has the shape shown in the figure.

The radius of the curved part of the wire is $R = 100 mm$ , the linear parts of the wire are very long

(permeability of vacuum,

μ_{0} = 1.257 \times 10^{- 6} H m^{- 1}

).

Accepted Answer

In cases (a) and (b), we have to find magnetic induction at the point $O$ due to three wires $1$ , $2$ and, $3$ in which $1$ and $3$ are semi-infinite straight wires and $2$ is a semi-circular wire. Current in the wire is $i$ , and the radius of the circular-shaped wire is $R$ .

Magnetic induction due to wires $1$ , $2$ and $3$ are $B_{1}$ , $B_{2}$ , and $B_{3}$ individually, then the net magnetic induction $B_{0}$ is equal to the vector sum of $B_{1}$ , $B_{2}$ , and $B_{3}$ .

(a) $\vec{B_{0}} = \vec{B_{1}} + \vec{B_{2}} + \vec{B_{3}}$

$= \frac{μ_{0}}{4 π} \frac{i}{R} (- \vec{k}) + \frac{μ_{0}}{4 π} \frac{i}{R} π (- \vec{i}) + \frac{μ_{0}}{4 π} \frac{i}{R} (- \vec{k})$

$= - \frac{μ_{0}}{4 π} \frac{i}{R} [2 \vec{k} + π \vec{i}]$

So, $|\vec{B_{0}}| = \frac{μ_{0}}{4 π} \frac{i}{R} \sqrt{π^{2} + 4} = 10^{- 7} \frac{8}{0.1} \times \sqrt{{(3.14)}^{2} + 4} = 0.30 μT$ .

(b) $\vec{B_{0}} = \vec{B_{1}} + \vec{B_{2}} + \vec{B_{3}}$
$= \frac{μ_{0}}{4 π} \frac{i}{R} (- \vec{k}) + \frac{μ_{0}}{4 π} \frac{i}{R} π (- \vec{i}) + \frac{μ_{0}}{4 π} \frac{i}{R} (- \vec{i})$
$= - \frac{μ_{0}}{4 π} \frac{i}{R} [\vec{k} + (π + 1) \vec{i}]$

So, $|\vec{B_{0}}| = \frac{μ_{0}}{4 π} \frac{i}{R} \sqrt{1 + (π + 1)^{2}} = 10^{- 7} \frac{8}{0.1} \times \sqrt{1 + (3.14 + 1)^{2}} = 0.34 μT$ .

(c) First, using the concept of division of current in a parallel combination of resistors, we calculate current in both sections of the circular wire. If the current in the lower section of the wire is $i_{2}$ , and the upper section of the wire is $i_{1}$ , then using the law of parallel resistances,

Here, using the law of parallel resistances,

$i_{1} + i_{2} = i$ and $\frac{i_{1}}{i_{2}} = \frac{1}{3}$

So, $\frac{i_{1}}{i_{2}} + 1 = \frac{1}{3} + 1$

$\frac{i_{1} + i_{2}}{i_{2}} = \frac{4}{3}$ .

Hence, $i_{2} = \frac{3}{4} i$ , and $i_{1} = \frac{1}{4} i$ .

Then, magnetic induction due to all the current-carrying wires, $\vec{B_{0}} = \frac{μ_{0}}{4 π} \frac{i}{R} (- \vec{k}) + \frac{μ_{0}}{4 π} \frac{i}{R} (- \vec{j}) + [\frac{μ_{0}}{4 π} (\frac{3 π}{2}) \frac{i_{1}}{R} (- i) + \frac{μ_{0}}{4 π} \frac{(\frac{π}{2}) i_{2}}{R} \vec{i}]$

$\vec{B_{0}} = - \frac{μ_{0}}{4 π} \frac{i}{R} (\vec{j} + \vec{k}) + 0$

Thus, $|\vec{B_{0}}| = \frac{μ_{0}}{4 π} \frac{\sqrt{2} i}{R} = 10^{- 7} \times \frac{8 \sqrt{2}}{0.1} = 0.11 μT$ .

Find the magnetic induction at the point $O$ , if the wire carrying a current $I = 8.0 A$ has the shape shown in the figure.

The radius of the curved part of the wire is $R = 100 mm$ , the linear parts of the wire are very long
(permeability of vacuum, $μ_{0} = 1.257 \times 10^{- 6} H m^{- 1}$ ).

Important Questions on Moving Charges and Magnetism

Learn and Prepare for any exam you want !

Find the magnetic induction at the point O, if the wire carrying a current I=8.0 A has the shape shown in the figure. The radius of the curved part of the wire is R=100 mm, the linear parts of the wire are very long (permeability of vacuum, μ0=1.257×10-6 H m-1).

Important Questions on Moving Charges and Magnetism

Learn and Prepare for any exam you want !

Find the magnetic induction at the point $O$ , if the wire carrying a current $I = 8.0 A$ has the shape shown in the figure.

The radius of the curved part of the wire is $R = 100 mm$ , the linear parts of the wire are very long
(permeability of vacuum, $μ_{0} = 1.257 \times 10^{- 6} H m^{- 1}$ ).