magnetic field
... A wire placed in magnetic field such that its length is perpendicular to the field which is of strength 10milliTesla. If the current in the wire is 15 milliamps and the magnitude of the force on the wire is 30x10-6 N, what is the length (in cm) of the wire in the magnetic field? ...
... A wire placed in magnetic field such that its length is perpendicular to the field which is of strength 10milliTesla. If the current in the wire is 15 milliamps and the magnitude of the force on the wire is 30x10-6 N, what is the length (in cm) of the wire in the magnetic field? ...
Solutions
... 9. Rail guns have been suggested for launching projectiles into space without chemical rockets and for ground-to-air antimissile weapons of war. A tabletop model rail gun consists of two long, parallel, horizontal rails 3.50 cm apart, bridged by a bar BD of mass 3.00 g. The bar is originally at res ...
... 9. Rail guns have been suggested for launching projectiles into space without chemical rockets and for ground-to-air antimissile weapons of war. A tabletop model rail gun consists of two long, parallel, horizontal rails 3.50 cm apart, bridged by a bar BD of mass 3.00 g. The bar is originally at res ...
Three dimensions Consider a point charge in three
... In this problem, the electric field from a distribution of charge in 3, 2, and 1 dimension has been found using Gauss's law. The most noteworthy feature of the three solutions is that in each case, there is a different relation of the field strength to the distance from the source of charge. In each ...
... In this problem, the electric field from a distribution of charge in 3, 2, and 1 dimension has been found using Gauss's law. The most noteworthy feature of the three solutions is that in each case, there is a different relation of the field strength to the distance from the source of charge. In each ...
magnetic field
... • There are two magnetic fields, H and B. In a vacuum they are indistinguishable, differing only by a multiplicative constant that depends on the physical units. Inside a material they are different. The term magnetic field is historically reserved for H while using other terms for B. Informally, th ...
... • There are two magnetic fields, H and B. In a vacuum they are indistinguishable, differing only by a multiplicative constant that depends on the physical units. Inside a material they are different. The term magnetic field is historically reserved for H while using other terms for B. Informally, th ...
When a current-carrying loop is placed in a
... Ex. 6 - A coil of wire has an area of 2.0 x 10-4 m2, consists of 100 loops, and contains a current of 0.045 A. The coil is placed in a uniform magnetic field of magnitude 0.15 T. (a) Determine the magnetic moment of the coil. (b) Find the maximum torque that the magnetic field can exert on the coil ...
... Ex. 6 - A coil of wire has an area of 2.0 x 10-4 m2, consists of 100 loops, and contains a current of 0.045 A. The coil is placed in a uniform magnetic field of magnitude 0.15 T. (a) Determine the magnetic moment of the coil. (b) Find the maximum torque that the magnetic field can exert on the coil ...
MAGNETISM SOLUTIONS
... perpendicular to a uniform magnetic field of 0.75 T. Find the time for the proton to make one complete circular orbit. 3A. (1) T = (2m) / (qB) (2) T = (2)(3.51 x 10-25 kg) / (3.24 x 10-11 C)(0.75 T) (3) T = 9.08 x 10-14 s 4. An electron moves through a region of crossed electric and magnetic field ...
... perpendicular to a uniform magnetic field of 0.75 T. Find the time for the proton to make one complete circular orbit. 3A. (1) T = (2m) / (qB) (2) T = (2)(3.51 x 10-25 kg) / (3.24 x 10-11 C)(0.75 T) (3) T = 9.08 x 10-14 s 4. An electron moves through a region of crossed electric and magnetic field ...
When a current-carrying loop is placed in a magnetic field
... Ex. 6 - A coil of wire has an area of 2.0 x 10-4 m2, consists of 100 loops, and contains a current of 0.045 A. The coil is placed in a uniform magnetic field of magnitude 0.15 T. (a) Determine the magnetic moment of the coil. (b) Find the maximum torque that the magnetic field can exert on the coil ...
... Ex. 6 - A coil of wire has an area of 2.0 x 10-4 m2, consists of 100 loops, and contains a current of 0.045 A. The coil is placed in a uniform magnetic field of magnitude 0.15 T. (a) Determine the magnetic moment of the coil. (b) Find the maximum torque that the magnetic field can exert on the coil ...
CBSE 2008 Physics Solved Paper XII
... Working At resonance the charged particle crosses the gap between the dees at just the right time so that the alternating voltage that is applied between the dees, accelerates it. A magnetic field acts perpendicular to the plane of the dees. An electric field is applied between the gap of the dees b ...
... Working At resonance the charged particle crosses the gap between the dees at just the right time so that the alternating voltage that is applied between the dees, accelerates it. A magnetic field acts perpendicular to the plane of the dees. An electric field is applied between the gap of the dees b ...
3.2
... The 32-ounce mixture must have a 25% gold content. The amount of gold content must be 25% of 32 ounces, or (0.25)(32) = 8 ounces. 0.16x + 0.28y = 8 Therefore, the system of equations is x + y = 32 0.16x + 0.28y = 8 Blitzer, Intermediate Algebra, 4e – Slide #40 ...
... The 32-ounce mixture must have a 25% gold content. The amount of gold content must be 25% of 32 ounces, or (0.25)(32) = 8 ounces. 0.16x + 0.28y = 8 Therefore, the system of equations is x + y = 32 0.16x + 0.28y = 8 Blitzer, Intermediate Algebra, 4e – Slide #40 ...
Magnetism - Northern Highlands
... magnetic fields of individual electrons in each atom cancel with each other. (Lead and diamond) The atoms in a piece of aluminum are randomly arranged, so the alignment of the north and south poles changes from one atom to the next making it paramagnetic. ...
... magnetic fields of individual electrons in each atom cancel with each other. (Lead and diamond) The atoms in a piece of aluminum are randomly arranged, so the alignment of the north and south poles changes from one atom to the next making it paramagnetic. ...
CP4 Solution
... For region II (r > R) we are taking a path form the central axis (r = 0) radially through regions I and regions II and so we need to use both functional forms for the electric field in the appropriate regions. The potential difference between any point lying on a circle of radius r > R and the centr ...
... For region II (r > R) we are taking a path form the central axis (r = 0) radially through regions I and regions II and so we need to use both functional forms for the electric field in the appropriate regions. The potential difference between any point lying on a circle of radius r > R and the centr ...