First slide

Magnetism

Question

What is the resultant magnetic field induction at the third corner, if two unlike poles, each of strength ‘m’ are placed at the two corners of an equilateral triangle of side ‘a’

Moderate

A

$B_{n e t} = \frac{μ_{0}}{4 π} \frac{m}{a^{2}}$
B

$B_{n e t} = 2 \frac{μ_{0}}{4 π} \frac{m}{a^{2}}$
C

$B_{n e t} = 3 (\frac{μ_{0}}{4 π} \frac{m}{a^{2}})$
D

$B_{n e t} = 6 (\frac{μ_{0}}{4 π} \frac{m}{a^{2}})$

Solution

Magnetic field induction at the point ‘C’ due to the North pole is away from its
$B_{1} = \frac{μ_{0}}{4 π} \frac{m}{a^{2}}$
Magnetic field induction at the point C due to the south pole is towards the pole
$B_{2} = \frac{μ_{0}}{4 π} \frac{m}{a^{2}}$
Applying parallelogram Law of Vectors
$B_{1} = B_{2} = \frac{μ_{0}}{4 π} \frac{m}{d^{2}} = B [L e t s a y] f r o m f i g u r e, θ = 120 °$
$∴ B_{net} = \sqrt{B^{2} + B^{2} + 2 B^{2} \cos 120 °}$
$\begin{array}{l} \Rightarrow B_{net} = \sqrt{B^{2} + B^{2} - B^{2}} \end{array}$
$\Rightarrow B_{net} = B = \frac{μ_{0}}{4 π} \frac{m}{d^{2}},$
$t h i s i s t h e m a g n i t u d e o f t h e r e s u l \tan t i n d u c t i o n a t t h e t h i r d c o r n e r w i t h t h e d i r e c t i o n b e i n g p a r a l l e l t o l i n e c o n n e c t i n g t h e n o r t h p o l e t o t h e s o u t h p o l e$

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