मराठी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान इयत्ता ११

Find the Dimensions of Magnetic Field B. the Relevant Equation Are F = Q E , F = Q V B , and B = μ 0 I 2 π a ; Where F is Force, Q is Charge, V is Speed, I is Current, and a is Distance. - Physics

Advertisements
Advertisements

प्रश्न

Find the dimensions of magnetic field B.
The relevant equation are \[F = qE, F = qvB, \text{ and }B = \frac{\mu_0 I}{2 \pi a};\]

where F is force, q is charge, v is speed, I is current, and a is distance.

बेरीज
Advertisements

उत्तर

Magnetic field,
\[B = \frac{F}{qv}\]
\[\text{ Here, } \left[ F \right] = {\left[ {MLT}^{- 2} \right]}, \left[ q \right] = {\left[ AT \right]} \text{ and } \left[ v \right] = {\left[ {LT}^{- 1} \right]} \]
\[\text{ So, dimension of magnetic field }, [B] = \frac{\left[ {MLT}^{- 2} \right]}{\left[ AT \right] \left[ {LT}^{- 1} \right]} = \left[ M L^0 T^{- 2} A^{- 1} \right]\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 1: Introduction to Physics - Exercise [पृष्ठ ९]

APPEARS IN

एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
पाठ 1 Introduction to Physics
Exercise | Q 3.2 | पृष्ठ ९

संबंधित प्रश्‍न

Some of the most profound statements on the nature of science have come from Albert Einstein, one of the greatest scientists of all time. What do you think did Einstein mean when he said : “The most incomprehensible thing about the world is that it is comprehensible”?


It is desirable that the standards of units be easily available, invariable, indestructible and easily reproducible. If we use foot of a person as a standard unit of length, which of the above features are present and which are not?


Find the dimensions of pressure.


Is it possible to add two vectors of unequal magnitudes and get zero? Is it possible to add three vectors of equal magnitudes and get zero?


Can you add three unit vectors to get a unit vector? Does your answer change if two unit vectors are along the coordinate axes?


Can you add two vectors representing physical quantities having different dimensions? Can you multiply two vectors representing physical quantities having different dimensions?


Let \[\vec{A} = 3 \vec{i} + 4 \vec{j}\]. Write a vector \[\vec{B}\] such that \[\vec{A} \neq \vec{B}\], but A = B.


A vector is not changed if


The x-component of the resultant of several vectors
(a) is equal to the sum of the x-components of the vectors of the vectors
(b) may be smaller than the sum of the magnitudes of the vectors
(c) may be greater than the sum of the magnitudes of the vectors
(d) may be equal to the sum of the magnitudes of the vectors.


The magnitude of the vector product of two vectors \[\left| \vec{A} \right|\] and \[\left| \vec{B} \right|\] may be

(a) greater than AB
(b) equal to AB
(c) less than AB
(d) equal to zero.


Two vectors have magnitudes 2 unit and 4 unit respectively. What should be the angle between them if the magnitude of the resultant is (a) 1 unit, (b) 5 unit and (c) 7 unit.


A spy report about a suspected car reads as follows. "The car moved 2.00 km towards east, made a perpendicular left turn, ran for 500 m, made a perpendicular right turn, ran for 4.00 km and stopped". Find the displacement of the car.


Suppose \[\vec{a}\] is a vector of magnitude 4.5 units due north. What is the vector (a) \[3 \vec{a}\], (b) \[- 4 \vec{a}\] ?


Draw a graph from the following data. Draw tangents at x = 2, 4, 6 and 8. Find the slopes of these tangents. Verify that the curve draw is y = 2x2 and the slope of tangent is \[\tan \theta = \frac{dy}{dx} = 4x\] 
\[\begin{array}x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ y & 2 & 8 & 18 & 32 & 50 & 72 & 98 & 128 & 162 & 200\end{array}\]


A curve is represented by y = sin x. If x is changed from \[\frac{\pi}{3}\text{ to }\frac{\pi}{3} + \frac{\pi}{100}\] , find approximately the change in y. 


Jupiter is at a distance of 824.7 million km from the Earth. Its angular diameter is measured to be 35.72˝. Calculate the diameter of Jupiter.


High speed moving particles are studied under


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×