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प्रश्न
The magnitude of the magnetic field due to a circular coil of radius R carrying a current I at an axial distance x from the centre is ______.
पर्याय
B = `(mu_0"IR"^3)/(2(x^2 + "R"^2)^(3/2))`
B = `(mu_0"IR"^2)/(2(x^2 + "R"^2)^(3/5))`
B = `(mu_0"IR"^2)/(2(x^2 + "R"^2)^(3/2))`
B = `(mu_0"IR"^2)/(3(x^2 + "R"^2)^(3/2))`
MCQ
रिकाम्या जागा भरा
उत्तर
The magnitude of the magnetic field due to a circular coil of radius R carrying a current I at an axial distance x from the centre is `underline("B" = (mu_0"IR"^2)/(2(x^2 + "R"^2)^(3/2)))`.
Explanation:
Magnetic field at a point located at a distance x from the center of the coil of radius R carrying current along its axis is B = `(mu_0"IR"^2)/(2(x^2 + "R"^2)^(3/2))`.
shaalaa.com
Magnetic Field on the Axis of a Circular Current Loop
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