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wire is bent and placed in a uniform magnetic field B. Current flows in the wire as shown in FIGURE 1 (a) A ЗА 0 cm 10 cm B-0.12 T 420 Q 20 cm FIGURE 1 Determine the magnitude and direction of the force that acts on each segment PQ, QR and RS [5 marks] The length of a 500 turns solenoid is 3 cm. Calculate the magnetic field inside the [1 marks (b) solenoid when it carries a 0.6 A current (c) A particle with a charge of 48 x 10-19 C enters a uniform magnetic field with a velocity of 2.0x 10 ms perpendicular to the field. Calculate the force that exerts 2 marks] (d) A 100 turns coil of mean cross-sectional area 15 cm2 carries 2.4 A current. on the particle if the magnitude of the field is 1.5 T Calculate the maximum torque on the coil when it is placed in a 0.8 T magnetic field 2 marks

Answer 1

A) The Lorentz force acting on the current carrying conductor in the magnetic filed is F = I B l sin theta For segment PQ, theta = 90 degree F_PQ = (3A) (0.12T) (0.10 m) sin 90 degree = 0.036 N For segment QR, since, wire QR is parallel to the direction of magnetic filed, then theta = 0 degree F_QR = 0 N For segment RS, theta = 42 degree F_RS = (3A) (0.12 T) (0.30m) sin 42 degree = 0.07226 N b) Magnetic filed inside the solenoid B = mu_0 pi I = mu_0 N/l I B = (4 pi times 10^-7 T. m/A) (500/3 times 10^-2 m) (0.6 A) B = 0.012566 T c) Force (F) = q V B sin theta F = (4.8 times 10^-19 c) (2 times 10^5 m/s) (1.5 T) sin 90 degree F = 1.44 times 10^-13 N d) Torque (T) = N I A B sin phivariant For maximum torque, phivariant = 90 degree