MHT CET · Physics · Magnetic Effects of Current
Two coils \(P\) and \(Q\) each of radius \(R\) carry currents I and \(\sqrt{8} \mathrm{I}\) respectively in same direction. Those coils are lying in perpendicular planes such that they have a common centre. The magnitude of the magnetic field at the common centre of the two coils is ( \(\mu_0=\) permeability of free space)
- A \(\frac{\mu_0 \mathrm{I}}{2 \mathrm{R}}\)
- B \(\frac{3 \mu_0 I}{2 R}\)
- C \(\frac{5 \mu_0 I}{2 R}\)
- D \(\frac{7 \mu_0 \mathrm{I}}{2 \mathrm{R}}\)
Answer & Solution
Correct Answer
(B) \(\frac{3 \mu_0 I}{2 R}\)
Step-by-step Solution
Detailed explanation
Magnetic field at the centre of P due to its current I is
\(\overrightarrow{\mathrm{B}_{\mathrm{p}}}=\frac{\mu_0 \mathrm{I}}{2 \mathrm{R}}...(i)\)
Magnetic field at the centre of Q due to its current \(\sqrt{8} 1\) is
\(\begin{aligned}
& \overrightarrow{\mathrm{B}_{\mathrm{Q}}} =\frac{\mu_0 \sqrt{8} \mathrm{I}}{2 \mathrm{R}}...(ii) \\
& \therefore \overline{\mathrm{~B}_{\text {net }}} \quad \ldots \text{[From(i) and (ii)]} \\
& =\sqrt{\mathrm{B}_{\mathrm{p}}^2+\mathrm{B}_{\mathrm{Q}}^2} \\
& =\sqrt{\left(\frac{\mu_0 I}{2 R}\right)^2+\left(\frac{\mu_0 \sqrt{8} I}{2 R}\right)^2} \\ & =\frac{3 \mu_0 I}{2 R}\end{aligned}\)
\(\overrightarrow{\mathrm{B}_{\mathrm{p}}}=\frac{\mu_0 \mathrm{I}}{2 \mathrm{R}}...(i)\)
Magnetic field at the centre of Q due to its current \(\sqrt{8} 1\) is
\(\begin{aligned}
& \overrightarrow{\mathrm{B}_{\mathrm{Q}}} =\frac{\mu_0 \sqrt{8} \mathrm{I}}{2 \mathrm{R}}...(ii) \\
& \therefore \overline{\mathrm{~B}_{\text {net }}} \quad \ldots \text{[From(i) and (ii)]} \\
& =\sqrt{\mathrm{B}_{\mathrm{p}}^2+\mathrm{B}_{\mathrm{Q}}^2} \\
& =\sqrt{\left(\frac{\mu_0 I}{2 R}\right)^2+\left(\frac{\mu_0 \sqrt{8} I}{2 R}\right)^2} \\ & =\frac{3 \mu_0 I}{2 R}\end{aligned}\)
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