A bar magnet of magnetic moment 5 Am 2 is placed in a uniform magnetic induction 3 × 1 0 − 5 T . If each pole of a magnet experiences a force of 2 ⋅ 5 × 1 0 − 4 N then the magnetic length of the magnet is

F = mB where m is pole strength. ∴ m = B F = 3 × 1 0 − 5 2.5 × 1 0 − 4 = 3 25 Am Magnetic moment M = mL ∴ L = m M = 25 5 × 3 = 0.6 m

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Physics - Magnetism Question with Solution | TestHub

PhysicsMagnetismMagnetic Dipole & MomentHard2 minPYQ_2020

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A bar magnet of magnetic moment $5 Am^{2}$ is placed in a uniform magnetic induction $3 \times 1 0^{- 5} T$ . If each pole of a magnet experiences a force of $2 \cdot 5 \times 1 0^{- 4} N$ then the magnetic length of the magnet is

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$F = mB$ where $m$ is pole strength. $∴ m = \frac{F}{B} = \frac{2.5 \times 1 0 ^{- 4}}{3 \times 1 0 ^{- 5}} = \frac{25}{3} Am$ Magnetic moment $M = mL$ $∴ L = \frac{M}{m} = \frac{5 \times 3}{25} = 0.6 m$

Stream:MHT_CETSubject:PhysicsTopic:MagnetismSubtopic:Magnetic Dipole & Moment

⏱ 2mℹ️ Source: PYQ_2020

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