Three rings, each having equal radius R , are placed mutually perpendicular to each other and each having its centre at the origin of the coordinate system. If a current I is flowing through each ring, then the magnitude of the magnetic field at the common centre is

Question

Three rings, each having equal radius R , are placed mutually perpendicular to each other and each having its centre at the origin of the coordinate system. If a current I is flowing through each ring, then the magnitude of the magnetic field at the common centre is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-aml2ahbollpbcsfk.jpg" /> (A) √3(μ 0 I/2 R) (B) zero (C) (√2 - 1)((μ )0 I/2 R) (D) (√3 - √2)((μ )0 I/2 R)

Tardigrade Admin · Accepted Answer

vecB=Bx hati+By hatj+Bz hatk ⇒ B=√Bx2+By2+Bz2 Since Bx=By=Bz so B=√3 B0 B=√3 (μ0 I/2 R)

Q. Three rings, each having equal radius $R$ , are placed mutually perpendicular to each other and each having its centre at the origin of the coordinate system. If a current $I$ is flowing through each ring, then the magnitude of the magnetic field at the common centre is

$3 \frac{μ _{0} I}{2 R}$

zero

$(2 - 1) \frac{( μ ) _{0} I}{2 R}$

$(3 - 2) \frac{( μ ) _{0} I}{2 R}$

$B = B_{x} \hat{i} + B_{y} \hat{j} + B_{z} \hat{k}$
$\Rightarrow B = B_{x}^{2} + B_{y}^{2} + B_{z}^{2}$
Since $B_{x} = B_{y} = B_{z}$ so $B = 3 B_{0}$
$B = 3 \frac{μ _{0} I}{2 R}$

Q. Three rings, each having equal radius R , are placed mutually perpendicular to each other and each having its centre at the origin of the coordinate system. If a current I is flowing through each ring, then the magnitude of the magnetic field at the common centre is

3​2Rμ0​I​

zero

(2​−1)2R(μ)0​I​

(3​−2​)2R(μ)0​I​

B=Bx​i^+By​j^​+Bz​k^ ⇒B=Bx2​+By2​+Bz2​​ Since Bx​=By​=Bz​ so B=3​B0​ B=3​2Rμ0​I​

Q. Three rings, each having equal radius $R$ , are placed mutually perpendicular to each other and each having its centre at the origin of the coordinate system. If a current $I$ is flowing through each ring, then the magnitude of the magnetic field at the common centre is

$3 \frac{μ _{0} I}{2 R}$

$(2 - 1) \frac{( μ ) _{0} I}{2 R}$

$(3 - 2) \frac{( μ ) _{0} I}{2 R}$

$B = B_{x} \hat{i} + B_{y} \hat{j} + B_{z} \hat{k}$
$\Rightarrow B = B_{x}^{2} + B_{y}^{2} + B_{z}^{2}$
Since $B_{x} = B_{y} = B_{z}$ so $B = 3 B_{0}$
$B = 3 \frac{μ _{0} I}{2 R}$