In an electric dipole there are two equal and opposite charges separated by a distance while in a magnetic dipole there are two equal and opposite poles (north and south – also called positive and negative- poles). The important difference in this context is that electric charges can be obtained as free positive and negative charges where as magnetic ‘charges’ (poles) can not be obtained as free north and south poles. They appear as dipoles only. Yet you will often be asked to caculate the pole strength of a magnetic dipole similar to the calculation of the charge in an electric dipole.
Let us consider the following questions (M.C.Q.):
(1) When an electric dipole of length 1cm is placed at an angle of 30º with an electric field of intensity 4×10^6 v/m, it experiences a torque of 2Nm. The charge on the dipole is
(a) 20μC
(b) 40μC
(c) 60μC
(d) 80μC
(e) 100μC
We have (with usual notation), torque τ = P×E = PEsinθ = P×4×10^6×½ = 2, from which P = 10^-6. But P = qL (L is the full length of the dipole), so that charge, q = P/L =10^-6/0.01 = 10^-4C = 100μC. The correct option therefore is (e).
(2) When a bar magnet of length 10cm is placed at an angle of 30º with a magnetic field of flux density 0.4 tesla, it experiences a torque of 0.6Nm. The pole strength of the magnet is (in Am)
(a) 40 (b) 30 (c) 20 (d) 10 (e) 5
This question is similar to the first question and the formulae are also similar. With usual notations, we have, torque τ = m×B = mBsinθ = m×0.4×½ = 0.6, from which m = 3Am^2. But m = pL (L is the full length of the magnet), so that the pole strength, p=m/L = 3/0.1 =30Am.
Let us consider the following questions (M.C.Q.):
(1) When an electric dipole of length 1cm is placed at an angle of 30º with an electric field of intensity 4×10^6 v/m, it experiences a torque of 2Nm. The charge on the dipole is
(a) 20μC
(b) 40μC
(c) 60μC
(d) 80μC
(e) 100μC
We have (with usual notation), torque τ = P×E = PEsinθ = P×4×10^6×½ = 2, from which P = 10^-6. But P = qL (L is the full length of the dipole), so that charge, q = P/L =10^-6/0.01 = 10^-4C = 100μC. The correct option therefore is (e).
(2) When a bar magnet of length 10cm is placed at an angle of 30º with a magnetic field of flux density 0.4 tesla, it experiences a torque of 0.6Nm. The pole strength of the magnet is (in Am)
(a) 40 (b) 30 (c) 20 (d) 10 (e) 5
This question is similar to the first question and the formulae are also similar. With usual notations, we have, torque τ = m×B = mBsinθ = m×0.4×½ = 0.6, from which m = 3Am^2. But m = pL (L is the full length of the magnet), so that the pole strength, p=m/L = 3/0.1 =30Am.
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