The length of a magnet of moment 5 Am2 is 14 cm. The magnetic induction at a point, equidistant from both poles is 3.5×10−5 wb/m2. the distance of the point from either pole is 100ncm where n=_____

Magnetic induction at distance ‘d’ from the centre of the magnetic of length 2lB=μ04πMd2+l232=μo4πMX3where x=d2+l2Given B=3.2×10−5wb/m2m=5 Am23.2×10−5=10−7×5X3⇒X3=5×10−73.2×10−5X=14m=14×100=25 cm=1004cm

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The length of a magnet of moment $5 A m^{2}$ is 14 cm. The magnetic induction at a point, equidistant from both poles is $3.5 \times 10^{- 5} w b / m^{2} .$ the distance of the point from either pole is $\frac{100}{n} c m$ where $n =_____$

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Detailed Solution

Magnetic induction at distance ‘d’ from the centre of the magnetic of length 2l

$B = \frac{μ_{0}}{4 π} \frac{M}{{(d^{2} + l^{2})}^{\frac{3}{2}}} = \frac{μ o}{4 π} \frac{M}{X^{3}}$ where $x = \sqrt{d^{2} + l^{2}}$

Given $B = 3.2 \times 10^{- 5}$ $w b / m^{2}$ $m = 5 A m^{2}$

$3.2 \times 10^{- 5} = 10^{- 7} \times \frac{5}{X^{3}} \Rightarrow X^{3} = \frac{5 \times 10^{- 7}}{3.2 \times 10^{- 5}}$

$X = \frac{1}{4} m = \frac{1}{4} \times 100 = 25 c m = \frac{100}{4} c m$

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