Magnetism

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The magnitude of magnetic field due to a dipole of magnetic moment 3.14 $A m^{2}$ at a point 1 m away from it in a direction making an angle of $60^{0}$ with the dipole axis is $\frac{\sqrt{7}}{2} . n \times 10^{- 8} T$ . Then the value of ‘n ’ is

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Solution

The magnitude of the field is $B = \frac{μ_{0}}{4 π} . \frac{M}{r^{3}} . \sqrt{1 + 3 \cos^{2} θ}$
$\Rightarrow B = (10^{- 7} T \frac{m}{A}) \frac{3.14 A m^{2}}{1 m} \sqrt{1 + 3 \cos^{2} 60^{0}}$
$\Rightarrow B = 3.14 \times 10^{- 7} . \sqrt{1 + \frac{3}{4}}$
$\Rightarrow B = 3.14 \times 10^{- 7} . \sqrt{\frac{7}{4}}$
$\Rightarrow B = (\frac{\sqrt{7}}{2}) (3.14) \times 10^{- 7}$
$\Rightarrow B = (\frac{\sqrt{7}}{2}) (31.4) \times 10^{- 8}$ T
$H e n c e, n = 31.4$

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Magnetism

Remember concepts with our Masterclasses.

The magnitude of magnetic field due to a dipole of magnetic moment 3.14 Am2 at a point 1 m away from it in a direction making an angle of 600 with the dipole axis is 72.n×10−8T. Then the value of ‘n ’ is

The magnitude of the field is B=μ04π.Mr3.1+3cos2θ ⇒B=10−7TmA3.14 Am21 m1+3cos2600 ⇒B=3.14×10−7.1+34⇒B=3.14×10−7.74 ⇒B=723.14×10−7 ⇒B=7231.4×10−8 THence, n=31.4

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The magnitude of magnetic field due to a dipole of magnetic moment 3.14 $A m^{2}$ at a point 1 m away from it in a direction making an angle of $60^{0}$ with the dipole axis is $\frac{\sqrt{7}}{2} . n \times 10^{- 8} T$ . Then the value of ‘n ’ is