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Physics NCERT Exemplar Solutions Class 11th Chapter Ten Magnetic Field due to a Current Element Physics Moving Charges and Magnetism Physics Class 12th

A wire $A$ , bent in the shape of an arc of a circle, carrying a current of $2 A$ and having radius $2 c m$ and another wire $B$ , also bent in the shape of arc of a circle, carrying a current of $3 A$ and having radius of $4 c m$ , are placed as shown in the figure. The ratio of the magnetic fields due to the wires $A$ and B at the common centre 0 is:

Option 1 -

$6 : 5$

Option 2 -

$2 : 5$

Option 3 -

$6 : 4$

Option 4 -

$4 : 6$

4 Views | Posted 3 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

3 months ago

Correct Option - 1

Detailed Solution:

$B_{A} = \frac{μ_{0} I θ}{4 π R}$

$\Rightarrow \frac{B_{A}}{B_{B}} = \frac{I_{A} θ_{A} R_{B}}{I_{B} θ_{B} R_{A}}$

$\Rightarrow \frac{2 (\frac{3 π}{2}) (4)}{3 [\frac{5 π}{3}] [2]}$

$\Rightarrow \frac{6}{5}$

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