Three rings, each having equal radius R and carrying a current , are placed mutually perpendicular to each

Question

Three rings, each having equal radius R and carrying a current $I$ , are placed mutually perpendicular to each other such that each has its centre at the origin of coordinate system. The magnitude of the magnetic field at the common centre is $\sqrt{k} (\frac{μ_{0} I}{2 R})$ . Then the value of $k$ is

Smart Achievers · Accepted Answer

(3)

$\vec{B} = \frac{μ_{0} I}{2 R} (\pm \hat{i}) + \frac{μ_{0} I}{2 R} (\pm \hat{j}) + \frac{μ_{0} I}{2 R} (\pm \hat{k})$

$\Rightarrow$ $| \vec{B} |$ $= \frac{μ_{0} I}{2 R} \sqrt{3}$

Solution

Ans.
(3)

$\vec{B} = \frac{μ_{0} I}{2 R} (\pm \hat{i}) + \frac{μ_{0} I}{2 R} (\pm \hat{j}) + \frac{μ_{0} I}{2 R} (\pm \hat{k})$

$\Rightarrow$ $| \vec{B} |$ $= \frac{μ_{0} I}{2 R} \sqrt{3}$

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Solution

Q. Three rings, each having equal radius R and carrying a current I, are placed mutually perpendicular to each other such that each has its centre at the origin of coordinate system. The magnitude of the magnetic field at the common centre is k(μ0I2R). Then the value of k is

Ans. (3) B→=μ0I2R(±i^)+μ0I2R(±j^)+μ0I2R(±k^) ⇒|B→|=μ0I2R3

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Ans.
(3)

$\vec{B} = \frac{μ_{0} I}{2 R} (\pm \hat{i}) + \frac{μ_{0} I}{2 R} (\pm \hat{j}) + \frac{μ_{0} I}{2 R} (\pm \hat{k})$

$\Rightarrow$ $| \vec{B} |$ $= \frac{μ_{0} I}{2 R} \sqrt{3}$