The magnetic induction field strength due to a short bar magnet of moment 3.6 A-m $^{2}$ at a distance of 0.2 m on the equatorial line is:

4.5 \times 10^{- 4} ⊤

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9 \times 10^{- 4} ⊤

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9 \times 10^{- 5} ⊤

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4.5 \times 10^{- 5} ⊤

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Solution

The correct option is D $4.5 \times 10^{- 5} ⊤$
Magnetic field at the equator is,
B(equitorial) $= \frac{- μ_{0} m}{4 π d^{3}}$
$= \frac{4 π \times 10^{- 7} \times 3.6}{4 π (0.2)^{3}}$
$= 4.5 \times 10^{- 5} T$

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The magnetic induction field strength due to a short bar magnet of moment 3.6 A-m2 at a distance of 0.2 m on the equatorial line is:

The correct option is D 4.5×10−5⊤Magnetic field at the equator is,B(equitorial) =−μ0m4πd3=4π×10−7×3.64π(0.2)3=4.5×10−5T

The magnetic induction field strength due to a short bar magnet of moment 3.6 A-m $^{2}$ at a distance of 0.2 m on the equatorial line is:

The correct option is D $4.5 \times 10^{- 5} ⊤$
Magnetic field at the equator is,
B(equitorial) $= \frac{- μ_{0} m}{4 π d^{3}}$
$= \frac{4 π \times 10^{- 7} \times 3.6}{4 π (0.2)^{3}}$
$= 4.5 \times 10^{- 5} T$