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A magnet of length 0.1 m and pole strength $ 10^ { -4 } A.m $ is kept in a magnetic field of 30 tesla at an angle of $30 ^ \circ $ . The couple acting on it is………. $ \times 10^ {-4 } $ Joule.
$ \tau = MB sin \theta $ $ = m ( 2l ) \times B sin \theta $ $ = 10^{-4} \times 0.1 \times 39 sin ^\circ $ $ = 1.5 \times 10^{-4} J $
In the case of bar magnet, lines of magnetic induction
In the case of a bar magnet, lines of magnetic induction (magnetic field lines) run continuously from the North pole to the South pole outside the magnet, and from the South pole to the North pole inside the magnet. This creates a closed loop, ensuring that the magnetic field lines are continuous.
A small bar magnet of moment M is placed in a uniform field of H. If magnet makes anangle of $ 30^ \circ $ with field, the torque acting on the magnet is
$ \tau = MB sin \theta $ $ = MH sin 30 ^ \circ $ $ = { MH \over 2 } $
The effective length of a magnet is 31.4 cm and its pole strength is 0.5 A.m. The magnetic moment, if it is bent in the form of a semicircle will be…….$Amp.m^2$.
$ 0.1 Amp m^2$
A bar magnet of length 10 cm and having the pole strength equal $ 00.1 \times 10^{-3} $ to is kept ina magnetic field having magnetic induction (B) equal to $4 \pi \times 10^ {-3 } $ tesla. It makes an angle of 300 with the direction of magnetic induction. The value of the torque acting on the magnet is ...................... Joule.
$ torque \tau = MH _H sin \theta $ $ = 2 \pi \times 10^{-7} J$
A small bar magnet has a magnetic moment $ 1.2 A.m^ 2$. The magnetic field at a distance 0.1 m on its axis will be…….tesla.
$ B = { \mu0 \over 4 \pi } { 2M \over d^3 } $ $ = 2.4 \times 10^{-4} tesla $
Force between two idential bur magnets whose centres are r meter apart is 4.8 N, when their axes are in the same line. If separation is increased to 2r, the force between them is reduced to
$ in magnetic dipole ,force \times {1 \over r^4 } $ $ { F_2 \over F_1 } = { r_1^4 \over r_2^4 } $ $ { F_2 \over 4.8 } = \left( { r_1 \over 2 r_1} \right)^4 $ $ = 0.3 Newton $
A magnet of magnetic moment $ 50 \uparrow A.m^2 $ is placed along the X-axis in a mag. Field $ \vec B = ( 0.5 \uparrow + 3.0 \hat J ) $ Tesla. The torque acting on the magnet is ……... N.m.
$ \tau = \vec M \times \vec B $ $ = 150 \hat K N.m$
A straight wire currying current I is turned into a circular Loop. If the magnitude of magnetic moment associated with it in MKs unit is M, the length of wire will be
Mag. moment of circular Loop carrying current is $ M = IA = I ( \pi R^2 ) = I \pi \left ( { L \over 2 \pi } \right) ^2 $ $ M = { IL^2 \over 4 \pi } \Rightarrow L = \sqrt { 4 \pi M \over I } $
A bar magnet is 10 cm long and is kept with its North (N) pole pointing North. A neutral point is formed at a distance of 15 cm from each pole. Given the horizontal component of earth's field to be 0.4 Gauss. The pole strength of the magnet is……. A.m.
$ L = 10 \times 10^{-2} m $ $ r = 15 \times 10^{-2} m $ $ OP = \sqrt { 225 -25 } $ $ = \sqrt 200 cm $ since, at theneutral point, magnetic field due to the magnetic equal to $B_H$ $ B_H = { \mu o \over 4 \pi } { M \over (OP^2 + AO^2 ) ^{3/2} }$ = 1.35 Amp. meter
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