Ncert Exercises & Solutions

5.12 A short bar magnet has a magnetic moment of 0.48 J T–1. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.

Magnetic moment of the bar magnet,

M = 0.48 J T^{- 1}

(a) Distance, d = 10 cm cm = 0.1 m

The magnetic field at distance d, from the centre of the magnet on the axis, is given by the relation:

B = \frac{μ_{0}}{4 π} \frac{2 M}{d^{3}}

Where,

μ_{0}

= Permeability of free space =

= 4 π \times 10^{- 7} Tm A^{- 1}

\begin{aligned} ∴ B & = \frac{4 π \times 10^{- 7} \times 2 \times 0.48}{4 π \times (0.1)^{3}} \\ = 0.96 \times 10^{- 4} T = 0.96 G \end{aligned}

The magnetic field is along the S-N direction.

(b) The magnetic field at a distance of 10 cm (i.e., d = 0.1 m) on

equatorial line of the magnet is given as:

\begin{array}{l} B = \frac{μ_{0}}{4 π} \frac{M}{d^{3}} \\ \Rightarrow B = \frac{4 π \times 10^{- 7} \times 0.48}{4 π \times (0.1)^{3}} \\ = 0.48 G \end{array}

The magnetic field is along the N - S direction.

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