Two short magnets of magnetic moment $1000 A m^{2}$ are placed as shown at the corners of a square of side $10 c m$ . The net magnetic induction at $P$ is

0.1 T

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0.2 T

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0.3 T

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0.4 T

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Solution

The correct option is A $0.1 T$
Magnetic Induction at P due to magnet M $B_{M} = \frac{μ_{0}}{4 π} \frac{2 M}{d^{3}}$ (at axial point)

Magnetic Induction at P due to magnet N $B_{N} = \frac{μ_{0}}{4 π} \frac{M}{d^{3}}$ (at equatorial point)

Net magnetic induction at P

$= B_{M} - B_{N}$

$= \frac{μ_{0}}{4 π d^{3}} M$

$= \frac{10^{- 7} \times 1000}{{(0.1)}^{3}}$

$= 0.1 T$

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Two short magnets of magnetic moment 1000 Am2 are placed as shown at the corners of a square of side 10 cm. The net magnetic induction at P is

The correct option is A 0.1 TMagnetic Induction at P due to magnet M BM=μ04π2Md3 (at axial point) Magnetic Induction at P due to magnet N BN=μ04πMd3 (at equatorial point) Net magnetic induction at P =BM−BN =μ04πd3M =10−7×1000(0.1)3 =0.1T

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