JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
Let \(B_1\) be the magnitude of magnetic field at center of a circular coil of radius \(R\) carrying current \(I\). Let \(B_2\) be the magnitude of magnetic field at an axial distance ' \(x\) ' from the center. For \(x: R=3: 4, \frac{B_2}{B_1}\) is :
- A \(4: 5\)
- B \(16: 25\)
- C \(64: 125\)
- D \(25: 16\)
Answer & Solution
Correct Answer
(C) \(64: 125\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{B}_1=\frac{\mu_0 \mathrm{i}}{2 \mathrm{R}} \qquad \mathrm{B}_2=\mathrm{B}_1 \sin ^3 \theta \\ & \therefore \frac{\mathrm{~B}_2}{\mathrm{~B}_1}=\sin ^3 \theta=\left(\frac{4}{5}\right)^3=\frac{64}{125}\end{aligned}\)
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